How can I print out all possible letter combinations a given phone number can represent

Question

I just tried for my first programming interview and one of the questions was to write a program that given a 7 digit telephone number  could print all possible combinations of letters that each number could represent   A second part of the question was like what about if this would have been a 12 digit international number   How would that effect your design   I don t have the code that I wrote in the interview  but I got the impression he wasn t happy with it  What is the best way to do this

User · Answer

static final String   keypad             ABC    DEF    GHI    JKL    MNO    PQRS    TUV    WXYZ       String   printAlphabet int num           if  num  gt   0  amp  amp  num  lt  10               String   retStr              if  num    0    num   1                   retStr   new String                      else                   retStr   new String keypad num  length                     for  int i   0   i  lt  keypad num  length    i                         retStr i    String valueOf keypad num  charAt i                                                return retStr                     String   nxtStr   printAlphabet num 10            int digit   num   10           String   curStr   null          if digit    0    digit    1               curStr   new String                  else               curStr   new String keypad digit  length                 for  int i   0  i  lt  keypad digit  length    i                     curStr i    String valueOf keypad digit  charAt i                                     String   result   new String curStr length   nxtStr length           int k 0           for  String cStr   curStr               for  String nStr   nxtStr                   result k      nStr   cStr                                  return result

User · Answer

The obvious solution is a function to map a digit to a list of keys  and then a function that would generate the possible combinations   The first is obvious  the second is more problematic because you have around 3 number of digits combinations  which can be a very large number   One way to do it is to look at each possibility for digit matching as a digit in a number  on  base 4  and implement something close to a counter  jumping over some instances  since there are usually less than 4 letters mappable to a digit    The more obvious solutions would be nested loops or recursion  which are both less elegant  but in my opinion valid   Another thing for which care must be taken is to avoid scalability issues  e g  keeping the possibilities in memory  etc   since we are talking about a lot of combinations   P S  Another interesting extension of the question would be localization

User · Answer

It is similar to a question called letter combinations of a phone number  here is my solution  It works for an arbitrary number of digits  so long as the result doesn t exceed the memory limit   import java util HashMap  public class Solution       public ArrayList lt String gt  letterCombinations String digits            ArrayList lt String gt  res   new ArrayList lt String gt             ArrayList lt String gt  preres   new ArrayList lt String gt             res add               for int i   0  i  lt  digits length    i                  String letters   map get digits charAt i                if  letters length      0                  continue              for String str   res                    for int j   0  j  lt  letters length    j                        preres add str   letters charAt j                              res   preres              preres   new ArrayList lt String gt                             return res             static final HashMap lt Character String gt  map   new HashMap lt Character String gt              put  1                put  2   abc            put  3   def            put  4   ghi            put  5   jkl            put  6   mno            put  7   pqrs            put  8   tuv            put  9   wxyz            put  0                     I m not sure how 12-digit international numbers affect the design   Edit  International numbers will also be handled

User · Answer

In Java using recursion   import java util LinkedList  import java util List   public class Main            Number-to-letter mappings in order from zero to nine     public static String mappings                   0      1      A    B    C      D    E    F      G    H    I              J    K    L      M    N    O      P    Q    R    S               T    U    V      W    X    Y    Z               public static void generateCombosHelper List lt String gt  combos               String prefix  String remaining               The current digit we are working with         int digit   Integer parseInt remaining substring 0  1             if  remaining length      1                   We have reached the last digit in the phone number  so add                 all possible prefix-digit combinations to the list             for  int i   0  i  lt  mappings digit  length  i                      combos add prefix   mappings digit  i                            else                  Recursively call this method with each possible new                 prefix and the remaining part of the phone number              for  int i   0  i  lt  mappings digit  length  i                      generateCombosHelper combos  prefix   mappings digit  i                            remaining substring 1                                       public static List lt String gt  generateCombos String phoneNumber               This will hold the final list of combinations         List lt String gt  combos   new LinkedList lt String gt                 Call the helper method with an empty prefix and the entire             phone number as the remaining part          generateCombosHelper combos      phoneNumber            return combos             public static void main String   args            String phone    3456789           List lt String gt  combos   generateCombos phone            for  String s   combos                System out println s

User · Answer

I ve implemented a cache which helped to reduce number of recursions  This cache will avoid scanning of letters that it had already done for previous combinations   For one instance it was going for 120k loops  after cache implementation it got reduced to 40k    private List lt String gt  generateWords String phoneNumber         List lt String gt  words   new LinkedList lt String gt         List lt String gt  wordsCache   new LinkedList lt String gt          this generatePossibleWords     phoneNumber  words  wordsCache        return words     private void generatePossibleWords String prefix  String remainder      List lt String gt  words  List lt String gt  wordsCache         int index   Integer parseInt remainder substring 0  1         for  int i   0  i  lt  phoneKeyMapper get index  size    i               String mappedChar   phoneKeyMapper get index  get i            if  prefix equals      amp  amp   wordsCache isEmpty                   for  String word   wordsCache                    words add mappedChar   word                           else                if  remainder length      1                    String stringToBeAdded   prefix   mappedChar                  words add stringToBeAdded                   wordsCache add stringToBeAdded substring 1                    LOGGER finest  adding stringToBeAdded       stringToBeAdded substring 1                   else                   generatePossibleWords prefix   mappedChar                  remainder substring 1   words  wordsCache                                    private void createPhoneNumberMapper          if  phoneKeyMapper    null         phoneKeyMapper   new ArrayList lt  gt         phoneKeyMapper add 0  new ArrayList lt String gt          phoneKeyMapper add 1  new ArrayList lt String gt           phoneKeyMapper add 2  new ArrayList lt String gt          phoneKeyMapper get 2  add  A        phoneKeyMapper get 2  add  B        phoneKeyMapper get 2  add  C         phoneKeyMapper add 3  new ArrayList lt String gt          phoneKeyMapper get 3  add  D        phoneKeyMapper get 3  add  E        phoneKeyMapper get 3  add  F         phoneKeyMapper add 4  new ArrayList lt String gt          phoneKeyMapper get 4  add  G        phoneKeyMapper get 4  add  H        phoneKeyMapper get 4  add  I         phoneKeyMapper add 5  new ArrayList lt String gt          phoneKeyMapper get 5  add  J        phoneKeyMapper get 5  add  K        phoneKeyMapper get 5  add  L         phoneKeyMapper add 6  new ArrayList lt String gt          phoneKeyMapper get 6  add  M        phoneKeyMapper get 6  add  N        phoneKeyMapper get 6  add  O         phoneKeyMapper add 7  new ArrayList lt String gt          phoneKeyMapper get 7  add  P        phoneKeyMapper get 7  add  Q        phoneKeyMapper get 7  add  R        phoneKeyMapper get 7  add  S         phoneKeyMapper add 8  new ArrayList lt String gt          phoneKeyMapper get 8  add  T        phoneKeyMapper get 8  add  U        phoneKeyMapper get 8  add  V         phoneKeyMapper add 9  new ArrayList lt String gt          phoneKeyMapper get 9  add  W        phoneKeyMapper get 9  add  X        phoneKeyMapper get 9  add  Y        phoneKeyMapper get 9  add  Z

User · Answer

public class Permutation          display all combination attached to a 3 digit number      public static void main String ar                  char data      new char       a   k   u                                      b   l   v                                      c   m   w                                      d   n   x                                      e   o   y                                      f   p   z                                      g   q   0                                      h   r   0                                      i   s   0                                      j   t   0               int num1  num2  num3 0          char tempdata      new char 3  3           StringBuilder number   new StringBuilder  324       a 3 digit number            copy data to a tempdata array-------------------         num1  Integer parseInt number substring 0 1            tempdata 0    data num1           num2  Integer parseInt number substring 1 2            tempdata 1    data num2           num3  Integer parseInt number substring 2 3            tempdata 2    data num3              display all combinations--------------------         char temp2     tempdata          char tempd  tempd2          int i i2  i3 0          for i 0 i lt 3 i                     tempd   temp2 0  i                for  i2 0 i2 lt 3 i2                      tempd2   temp2 1  i2                    for i3 0 i3 lt 3 i3                          System out print tempd                        System out print tempd2                        System out print temp2 2  i3                         System out println                        for i3                 for i2                      end of class

User · Answer

In Python  iterative     digit map          2    abc        3    def        4    ghi        5    jkl        6    mno        7    pqrs        8    tuv        9    wxyz      def word numbers input     input   str input    ret          for char in input      letters   digit map get char          ret    prefix letter for prefix in ret for letter in letters    return ret   ret is a list of results so far  initially it is populated with one item  the empty string  Then  for each character in the input string  it looks up the list of letters that match it from the dict defined at the top  It then replaces the list ret with the every combination of existing prefix and possible letter

User · Answer

I am rather a newbie so please correct me wherever I am wrong   First thing is looking into space  amp  time complexity  Which is really bad since it s factorial so for factorial 7    5040 any recursive algorithm would do  But for factorial 12     4   10 8 which can cause stack overflow in recursive solution   So I would not attempt a recursive algorithm  Looping solution is very straight forward using  Next Permutation     So I would create and array  0  1  2  3  4  5  and generate all permutation and while printing replace them with respective characters eg  0 A  5 F  Next Perm algorithm works as follows  eg Given  1 3 5 4 next permutation is 1 4 3 5  Steps for finding next perm    From right to left  find first decreasing number  eg 3 From left to right  find lowest number bigger than 3 eg  4 Swap these numbers as reverse the subset  1 4 5 3 reverse subset 1 4 3 5   Using Next permutation   or rotation  you generate specific subset of permutations  say you want to show 1000 permutations starting from a particular phone number  This can save you from having all numbers in memory  If I store numbers as 4 byte integers  10 9 bytes   1 GB

User · Answer

This approach uses R and is based on first converting the dictionary to its corresponding digit representation  then using this as a look-up   The conversion only takes 1 second on my machine  converting from the native Unix dictionary of about 100 000 words   and typical look-ups of up to 100 different digit inputs take a total of  1 seconds   library data table   example dictionary dict orig   tolower readLines   usr share dict american-english      split each word into its constituent letters  words shorter than the longest padded with    for simpler retrieval dictDT   setDT tstrsplit dict orig  split       fill          lookup table for conversion  NB  the following are found in the dictionary and would need    to be handled separately -- ignoring here     accents should just be appended to     matches for unaccented version          c                                                                                                                      lookup   data table num   c rep  2   3   rep  3   3   rep  4   3                               rep  5   3   rep  6   3   rep  7   4                               rep  8   3   rep  9   4                        let   letters    using the lookup table  convert to numeric for  col in names dictDT       dictDT lookup   col     i num  on   setNames  let   col       back to character vector dict num   do call paste0  dictDT    sort both for faster vector search idx   order dict num  dict num   dict num idx  dict orig   dict orig idx   possibilities   function input  dict orig dict num    input    sample output  possibilities  269      1   amy   bmw   cox   coy   any   bow   box   boy   cow   cox   coy  possibilities  22737      1   acres   bards   barer   bares   barfs   baser   bases   caper     9   capes   cards   cares   cases

User · Answer

An R solution using nested loops     Create phone pad two  lt - c  A    B    C   three  lt - c  D    E    F   four  lt - c  G    H    I   five  lt - c  J    K    L   six  lt - c  M    N    O    P   seven  lt - c  Q    R    S   eight  lt - c  T    U    V   nine  lt - c  W    X    Y    Z      Choose three numbers number 1  lt - two number 2  lt - three number 3  lt - six    create an object to save the combinations to combinations  lt - NULL    Loop through the letters in number 1 for i in number 1          Loop through the letters in number 2     for j in number 2              Loop through the letters in number 3         for k in number 3                      Add each of the letters to the combinations object                 combinations  lt - c combinations  paste0 i  j  k                            Print all of the possible combinations combinations   I posted another  more bizarre R solution using more loops and sampling on my blog

User · Answer

In numeric keyboards  texts and numbers are placed on the same key  For example 2 has    ABC    if we wanted to write anything starting with    A    we need to type key 2 once  If we wanted to type    B     press key 2 twice and thrice for typing    C     below is picture of such keypad   keypad http   d2o58evtke57tz cloudfront net wp-content uploads phoneKeyboard png  Given a keypad as shown in diagram  and a n digit number  list all words which are possible by pressing these numbers   For example if input number is 234  possible words which can be formed are  Alphabetical order   adg adh adi aeg aeh aei afg afh afi bdg bdh bdi beg beh bei bfg bfh bfi cdg cdh cdi ceg ceh cei cfg cfh cfi  Let   s do some calculations first  How many words are possible with seven digits with each digit representing n letters  For first digit we have at most four choices  and for each choice for first letter  we have at most four choices for second digit and so on  So it   s simple maths it will be O 4 n   Since keys 0 and 1 don   t have any corresponding alphabet and many characters have 3 characters  4 n would be the upper bound of number of words and not the minimum words   Now let   s do some examples   For number above 234  Do you see any pattern  Yes  we notice that the last character always either G H or I and whenever it resets its value from I to G  the digit at the left of it gets changed  Similarly whenever the second last alphabet resets its value  the third last alphabet gets changes and so on  First character resets only once when we have generated all words  This can be looked from other end also  That is to say whenever character at position i changes  character at position i 1 goes through all possible characters and it creates ripple effect till we reach at end  Since 0 and 1 don   t have any characters associated with them  we should break as there will no iteration for these digits   Let   s take the second approach as it will be easy to implement it using recursion  We go till the end and come back one by one  Perfect condition for recursion  let   s search for base case  When we reach at the last character  we print the word with all possible characters for last digit and return  Simple base case When we reach at the last character  we print the word with all possible characters for last digit and return  Simple base case   Following is C implementation of recursive approach to print all possible word corresponding to a n digit input number  Note that input number is represented as an array to simplify the code    include  lt stdio h gt   include  lt string h gt      hashTable i  stores all characters that correspond to digit i in phone const char hashTable 10  5              abc    def    ghi    jkl                               mno    pqrs    tuv    wxyz        A recursive function to print all possible words that can be obtained    by input number   of size n   The output words are one by one stored    in output   void  printWordsUtil int number    int curr digit  char output    int n           Base case  if current output word is prepared int i  if  curr digit    n        printf   s    output       return         Try all 3 possible characters for current digir in number      and recur for remaining digits for  i 0  i lt strlen hashTable number curr digit     i          output curr digit    hashTable number curr digit   i       printWordsUtil number  curr digit 1  output  n       if  number curr digit     0    number curr digit     1          return          A wrapper over printWordsUtil     It creates an output array and    calls printWordsUtil   void printWords int number    int n    char result n 1   result n     0   printWordsUtil number  0  result  n        Driver program int main void    int number      2  3  4   int n   sizeof number  sizeof number 0    printWords number  n   return 0      Output   adg adh adi aeg aeh aei afg afh afi bdg bdh bdi beg beh bei bfg bfh bfi cdg cdh cdi ceg ceh cei cfg cfh cfi   Time Complexity   Time complexity of above code is O 4 n  where n is number of digits in input number   References   http   www flipkart com programming-interviews-exposed-secrets-landing-your-next-job-3rd p itmdxghumef3sdjn pid 9788126539116 amp affid sandeepgfg

User · Answer

This is the C  port of this answer  Code    public class LetterCombinations       private static readonly Dictionary lt string  string gt  Representations   new Dictionary lt string  string gt                 quot 2 quot    quot abc quot              quot 3 quot    quot def quot              quot 4 quot    quot ghi quot              quot 5 quot    quot jkl quot              quot 6 quot    quot mno quot              quot 7 quot    quot pqrs quot              quot 8 quot    quot tuv quot              quot 9 quot    quot wxyz quot                 public static List lt string gt  FromPhoneNumber string phoneNumber                var result   new List lt string gt    string Empty                go through each number in the phone         for  int i   0  i  lt  phoneNumber Length  i                          var pre   new List lt string gt                 foreach  var str in result                                var letters   Representations phoneNumber i  ToString                        go through each representation of the number                 for  int j   0  j  lt  letters Length  j                                          pre Add str   letters j                                                result   pre                     return result           Unit Tests public class UnitTest        TestMethod      public void One Digit Yields Three Representations                 var sut    quot 2 quot            var expected   new List lt string gt    quot a quot    quot b quot    quot c quot             var actualResults   LetterCombinations FromPhoneNumber sut            CollectionAssert AreEqual expected  actualResults               TestMethod      public void Two Digits Yield Nine Representations                 var sut    quot 22 quot            var expected   new List lt string gt     quot aa quot    quot ab quot    quot ac quot    quot ba quot    quot bb quot    quot bc quot    quot ca quot    quot cb quot    quot cc quot             var actualResults   LetterCombinations FromPhoneNumber sut            CollectionAssert AreEqual expected  actualResults               TestMethod      public void Three Digits Yield ThirtyNine Representations                 var sut    quot 222 quot            var actualResults   LetterCombinations FromPhoneNumber sut            var possibleCombinations   Math Pow 3 3     27          Assert AreEqual possibleCombinations  actualResults Count

User · Answer

One option in Objective-C     -  NSArray   lettersForNumber  NSNumber   number       switch   number intValue             case 2              return     A    B    C            case 3              return     D    E    F            case 4              return     G    H    I            case 5              return     J    K    L            case 6              return     M    N    O            case 7              return     P    Q    R    S            case 8              return     T    U    V            case 9              return     W    X    Y    Z            default              return nil           -  NSArray   letterCombinationsForNumbers  NSArray   numbers       NSMutableArray  combinations     NSMutableArray alloc  initWithObjects      nil        for  NSNumber  number in numbers            NSArray  lettersNumber    self lettersForNumber number              Ignore numbers that don t have associated letters         if  lettersNumber count    0                continue                     NSMutableArray  currentCombinations    combinations mutableCopy           combinations     NSMutableArray alloc  init            for  NSString  letter in lettersNumber                 for  NSString  letterInResult in currentCombinations                     NSString  newString    NSString stringWithFormat          letterInResult  letter                    combinations addObject newString                                      return combinations

User · Answer

Use a list L where L i    the symbols that digit i can represent   L 1           for example  L 2    a b c  Etc   Then you can do something like this  pseudo-C    void f int k  int st        if   k  gt  numberOfDigits           print contents of st        return         for each character c in L Digit At Position k          st k    c      f k   1  st           Assuming each list contains 3 characters  we have 3 7 possibilities for 7 digits and 3 12 for 12  which isn t that many  If you need all combinations  I don t see a much better way  You can avoid recursion and whatnot  but you re not going to get something a lot faster than this no matter what

User · Answer

Oracle SQL  Usable with any phone number length and can easily support localization   CREATE TABLE digit character map  digit number 1   character varchar2 1     SELECT replace permutations         AS permutations FROM  SELECT sys connect by path map CHARACTER      AS permutations  LEVEL AS lvl       FROM digit character map map        START WITH map digit   substr  12345  1 1        CONNECT BY   digit   substr  12345  LEVEL 1   WHERE lvl   length  12345

User · Answer

This is a recursive algorithm in C  11    include  lt iostream gt   include  lt array gt   include  lt list gt   std  array lt std  string  10 gt  pm          0    1    ABC    DEF    GHI    JKL    MNO    PQRS    TUV    WXYZ      void generate mnemonic const std  string amp  numbers  size t i  std  string amp  m      std  list lt std  string gt  amp  mnemonics           Base case     if  numbers size      i            mnemonics push back m           return             for  char c   pm numbers i  -  0              m i    c          generate mnemonic numbers  i   1  m  mnemonics            std  list lt std  string gt  phone number mnemonics const std  string amp  numbers        std  list lt std  string gt  mnemonics      std  string m numbers size    0       generate mnemonic numbers  0  m  mnemonics       return mnemonics     int main         std  list lt std  string gt  result   phone number mnemonics  2276696        for  const std  string amp  s   result            std  cout  lt  lt  s  lt  lt  std  endl            return 0

User · Answer

I rewrote the latest answer to this  referred above    from C to Java   I also included the support for 0 and 1  as 0 and 1  because numbers such as 555-5055 weren t working at all with the above code     Here it is  Some comments are preserved    public static void printPhoneWords int   number        char   output   new char number length       printWordsUtil number 0 output      static String   phoneKeys  new String    0    1    ABC    DEF    GHI    JKL                   MNO    PQRS    TUV    WXYZ    private static void printWordsUtil int   number  int curDigIndex  char   output           Base case  if current output word is done     if  curDigIndex    output length            System out print String valueOf output                   return                  Try all 3-4 possible characters for the current digit in number            and recurse for the remaining digits      char curPhoneKey     phoneKeys number curDigIndex   toCharArray        for  int i   0  i lt  curPhoneKey length   i              output curDigIndex    curPhoneKey i           printWordsUtil number  curDigIndex 1  output           if  number curDigIndex   lt   1     for 0 or 1             return           public static void main String   args        int number      2  3  4       printPhoneWords number       System out println

User · Answer

Simple Java implementation without any input error checking      expects all digits as input       public class PhoneSpeller        private static final char      letters   new char                    0                  1                  A    B    C                  D    E    F                  G    H    I                  J    K    L                  M    N    O                  P    Q    R    S                  T    U    V                  W    X    Y    Z               public static void main String   args                if  args length    1                        System out println  Please run again with your phone number  no dashes                 System exit 0                     recursive phoneSpell args 0   0  new ArrayList lt String gt                   private static void recursive phoneSpell String arg  int index  List lt String gt  results                if  index    arg length                          printResults results               return                    int num   Integer parseInt arg charAt index                if  index  0                        for  int j   0  j lt  letters num  length  j                                  results add  letters num  j                                 recursive phoneSpell arg  index 1  results                     else                       List lt String gt  combos   new ArrayList lt String gt                 for  int j   0  j lt  letters num  length  j                                   for  String result   results                                          combos add result  letters num  j                                                 recursive phoneSpell arg  index 1  combos                        private static void printResults List lt String gt  results                for  String result   results                        System out println result

User · Answer

During a job interview I received this question regarding creating an array of and printing out all possible letter combinations of a phone number  During the interview I received a whisper from my interviewer about recursion and how it can t be done using loops  Very unnatural for me to receive input from another programmer  I trusted his advice instead of my own tuition and proceeded to write up a sloppy recursion mess  It didn t go so well  Before receiving input  as I had never received this problem before  my brain was calculating up the underlying replicable mathematical formula  Whispers under my breath   can t just multiply by three  some of them are four  as my mind started racing against a short clock thinking about one answer while beginning to write another  It was not until the end of the week I received my call to let me know the sad news  It was later that night I decided to see if it really is a recursion problem  Turns out for me it isn t   My solution below is coded in PHP  is non-recursive  works with any length of input and I ve even included some comments to help describe what my variables mean  Its my official and unchanging final answer to this question  I hope this helps somebody else in the future  please feel free to translate this into other languages   Me method involves calculating the total number of combinations and creating a buffer large enough to hold every byte  The length of my buffer is the same length you would expect to receive from a working recursive algorithm  I make an array that contains the groups of letters that represent each number  My method primarily works because your combo total will always be divisible by the number of letters in the current group  If you can imagine in your head a vertical list of the output of this algorithm  or see the list vertically as opposed to a horizontally written list of the combinations  my algorithm will begin to make sense  This algorithm allows us to loop through each byte vertically from top to bottom starting from the left instead of horizontally left to right  and populate each byte individually without requiring recursion  I use the buffer as though its a byte array to save time and energy     NON-RECURSIVE  ITERATIVE PHP   lt  php     Display all possible combinations of letters for each number    input    23456789                             if  isset  input 0        die  Nothing to see here       phone letters   array      2   gt  array  a    b    c        3   gt  array  d    e    f        4   gt  array  g    h    i        5   gt  array  j    k    l        6   gt  array  m    n    o        7   gt  array  p    q    r    s        8   gt  array  t    u    v        9   gt  array  w    x    y    z        groups   array     combos total   1   l   strlen  input   for  i   0   i  lt   l     i         groups      phone letters  input  i         combos total    count  phone letters  input  i        count    combos total   count  groups 0     combos   array fill 0   combos total  str repeat chr 0   strlen  input      for       group   0                         Index for the current group of letters       groups count   count  groups      Total number of letter groups       letters   count  groups 0         Total number of letters in the current group       width    combos total             Number of bytes to repeat the current letter       repeat   1                        Total number of times the group will repeat                for  byte   0   width     letters   r   0   r  lt   repeat     r          for  l   0   l  lt   letters     l              for  w   0   w  lt   width     w                   combos  byte     group     groups  group   l        if    group  lt   groups count             repeat     letters           letters   count  groups  group              else         break                                if is array  combos         print r  combos       echo  Total combos    count  combos     n     else     echo  No combos      n   echo  Expected combos      combos total    n       NON-RECURSIVE  NON-ITERATIVE  NO-BUFFER  THREAD FRIENDLY  MATH ONLY   NON-RECURSIVE  ITERATIVE PHP OBJECT USING MULTI-BASE BIG ENDIAN  While I was working on the new house the other day  I had a realisation that I could use the mathematics from my first function coupled with my knowledge of base to treat all possible letter combinations of my input as a multi-dimensional number   My object provides the functions necessary to store a prefered combination into a database  convert letters and numbers if it were required  pick out a combination from anywhere in the combination space using a prefered combination or byte and it all plays very well with multi-threading   That being said  using my object I can provide a second answer that uses base  no buffer  no iterations and most importantly no recursion    lt  php class phone       public static  letters   array          2   gt  array  a    b    c            3   gt  array  d    e    f            4   gt  array  g    h    i            5   gt  array  j    k    l            6   gt  array  m    n    o            7   gt  array  p    q    r    s            8   gt  array  t    u    v            9   gt  array  w    x    y    z                  Convert a letter into its respective number      public static function letter number  letter            if  isset  letter 0      isset  letter 1                return false           for  i   2   i  lt  10     i              if in array  letter  phone   letters  i   true                   return  i           return false                Convert letters into their respective numbers      public static function letters numbers  letters            if  isset  letters 0                return false            length   strlen  letters            numbers   str repeat chr 0    length           for  i   0   i  lt   length     i                for  j   2   j  lt  10     j                    if in array  letters  i   phone   letters  j   true                          numbers  i     j                      break                                                    return  numbers                Calculate the maximum number of combinations that could occur within a particular input size      public static function combination size  groups            return  groups  lt   0   false   pow 4   groups                 Calculate the minimum bytes reqired to store a group using the current input      public static function combination bytes  groups            if  groups  lt   0              return false            size    groups   4           bytes   0          while  groups     0                 groups  gt  gt  8                 bytes                    return  bytes             public  input           public  input len   0      public  combinations total   0      public  combinations length   0       private  iterations   array        private  branches   array         function   construct  number            if  isset  number 0                return false            this- gt input    number           input len   strlen  number             combinations total   1          for  i   0   i  lt   input len     i                 combinations total    count phone   letters  number  i                         this- gt input len    input len           this- gt combinations total    combinations total           this- gt combinations length    combinations total    input len           for  i   0   i  lt   input len     i                 this- gt branches      this- gt combination branches  i                this- gt iterations      this- gt combination iteration  i                           Calculate a particular combination in the combination space and return the details of that combination      public function combination  combination             position    combination    this- gt input len          if  position  lt  0     position  gt    this- gt combinations length                  return false            group    position    this- gt input len           first    position -  group           last    first    this- gt input len - 1           combination   floor   last   1     this- gt input len  - 1            bytes   str repeat chr 0    this- gt input len           for  i   0   i  lt   this- gt input len     i               bytes  i    phone   letters  this- gt input  i     combination    this- gt branches  i     count phone   letters  this- gt input  i               return array               combination    gt   combination               branches    gt   this- gt branches  group                iterations    gt   this- gt iterations  group                first    gt   first               last    gt   last               bytes    gt   bytes                          Calculate a particular byte in the combination space and return the details of that byte      public function combination position  position            if  position  lt  0     position  gt    this- gt combinations length                  return false            group    position    this- gt input len           group count   count phone   letters  this- gt input  group              first    position -  group           last    first    this- gt input len - 1           combination   floor   last   1     this- gt input len  - 1           index     combination    this- gt branches  group      group count            bytes   str repeat chr 0    this- gt input len           for  i   0   i  lt   this- gt input len     i               bytes  i    phone   letters  this- gt input  i     combination    this- gt branches  i     count phone   letters  this- gt input  i               return array               position    gt   position               combination    gt   combination - 1               group    gt   group               group count    gt   group count               group index    gt   index               number    gt   this- gt input  group                branches    gt   this- gt branches  group                iterations    gt   this- gt iterations  group                first    gt   first               last    gt   last               byte    gt  phone   letters  this- gt input  group    index                bytes    gt   bytes                          Convert letters into a combination number using Multi-Base Big Endian      public function combination letters  letters            if  isset  letters  this- gt input len - 1      isset  letters  this- gt input len                return false            combination   0          for  byte   0   byte  lt   this- gt input len     byte                 base   count phone   letters  this- gt input  byte                  found   false              for  value   0   value  lt   base     value                    if  letters  byte      phone   letters  this- gt input  byte    value                          combination     value    this- gt branches  byte                        found   true                      break                                              if  found     false                  return false                    return  combination                Calculate the number of branches after a particular group      public function combination branches  group            if  group  gt   0  amp  amp     group  lt   this- gt input len                 branches   1              for  i    group   i  lt   this- gt input len     i                   branches    count phone   letters  this- gt input  i                 return  branches     1   1    branches                    elseif  group      this- gt input len              return 1          else         return false                   Calculate the number of iterations before a particular group      public function combination iteration  group            if  group  gt  0  amp  amp   group  lt   this- gt input len                 iterations   1              for  i    group - 1   i  gt   0  -- i                   iterations    count phone   letters  this- gt input  i                 return  iterations                    elseif  group     0              return 1          else             return false                Iterations  Outputs array of all combinations using a buffer      public function combinations              count    this- gt combinations total   count phone   letters  this- gt input 0              combinations   array fill 0   this- gt combinations total  str repeat chr 0    this- gt input len            for               group   0                                             Index for the current group of letters               groups count    this- gt input len                       Total number of letter groups               letters   count phone   letters  this- gt input 0        Total number of letters in the current group               width    this- gt combinations total                             Number of bytes to repeat the current letter               repeat   1                                            Total number of times the group will repeat                                        for  byte   0   width     letters   r   0   r  lt   repeat     r                  for  l   0   l  lt   letters     l                      for  w   0   w  lt   width     w                           combinations  byte     group    phone   letters  this- gt input  group    l                if    group  lt   groups count                     repeat     letters                   letters   count phone   letters  this- gt input  group                               else                 break                    return  combinations                                     phone   new phone  23456789    print r  phone     print r  phone- gt combinations     for  i   0   i  lt   phone- gt combinations total     i        echo  i         phone- gt combination  i   bytes      n

User · Answer

Here is the working code for the same   its a recursion on each step checking possibility of each combinations on occurrence of same digit in a series    I am not sure if there is any better solution then this from complexity point of view       Please let me know for any issues      import java util ArrayList    public class phonenumbers                    param args             public static String mappings                   0      1      A    B    C      D    E    F      G    H    I              J    K    L      M    N    O      P    Q    R    S               T    U    V      W    X    Y    Z               public static void main String   args               TODO Auto-generated method stub          String phone    3333456789           ArrayList lt String gt  list  generateAllnums phone    0              private static ArrayList lt String gt  generateAllnums String phone String sofar int j               TODO Auto-generated method stub            System out println phone           if phone isEmpty                 System out println sofar               for int k1 0 k1 lt sofar length   k1                     int m sofar toLowerCase   charAt k1 -48                  if m  -16                      continue                  int i k1                  while true  amp  amp  i lt sofar length  -2                       if sofar charAt i 1                                break                      else if sofar charAt i 1   sofar charAt k1                            i                         else                          break                                                           i i-k1                    System out print       m         i                         System out print mappings m  i 3                    k1 k1 i                            System out println                 return null                    int num  phone charAt j           int k 0          for int i j 1 i lt phone length   i                 if phone charAt i   num                   k                                     if k  0                int p 0              ArrayList lt String gt  list2  generateAllnums phone substring p 1   sofar phone charAt p        0               ArrayList lt String gt  list3  generateAllnums phone substring p 1   sofar phone charAt p   0                      else              ArrayList lt String gt  list1  generateAllnums phone substring 1   sofar phone charAt 0   0                     return null

User · Answer

I tried it in ruby  and came up with a different way of doing  it s probably not efficient  like time and space O    at this point  but I like it because it uses Ruby s builtin Array product method  What do you think   EDIT  I see a very similar solution in Python above  but I hadn t seen it when I added my answer  def phone to abc phone     phone abc          0    1    abc    def    ghi        jkl    mno    pqrs    tuv    wxyz         phone map   phone chars map    x  phone abc x to i  chars     result   phone map 0    for i in 1  phone map size-1     result   result product phone map i     end   result each    x      puts    x join        end  phone to abc  86352

User · Answer

You find source  Scala  here and an working applet here   Since 0 and 1 aren t matched to characters  they build natural breakpoints in numbers  But they don t occur in every number  except 0 at the beginning   Longer numbers like  49567892345 from 9 digits starting  can lead to OutOfMemoryErrors  So it would be better to split a number into groups like    01723 5864 0172 35864   to see  if you can make sense from the shorter parts  I wrote such a program  and tested some numbers from my friends  but found rarely combinations of shorter words  which could be checked in a dictionary for matching  not to mention single  long words    So my decision was to only support searching  no full automation  by displaying possible combinations  encouraging splitting the number by hand  maybe multiple time    So I found  -RAD JUNG   -bycicle boy     If you accept misspellings  abbreviations  foreign words  numbers as words  numbers in words  and names  your chance to find a solution is much better  than without fiddling around    246848   gt  2hot4u  too hot for you   466368   gt  goodn8  good night   1325     gt  1FCK    Football club  53517    gt  JDK17   Java Developer Kit    are things a human might observe - to make an algorithm find such things is rather hard

User · Answer

include  lt sstream gt   include  lt map gt   include  lt vector gt   map lt  int  string gt  keyMap   void MakeCombinations  string first  string joinThis   vector lt string gt  amp  eachResult         if   first size             return       int length   joinThis length        vector lt string gt  result       while  length                 string each          char firstCharacter   first at 0           each    firstCharacter          each    joinThis length -1           length--           result push back each                   first   first substr 1        vector lt string gt   iterator begin   result begin            vector lt string gt   iterator end   result end        while  begin    end                eachResult push back   begin           begin               return MakeCombinations  first  joinThis  eachResult       void ProduceCombinations  int inNumber  vector lt string gt  amp  result        vector lt string gt  inputUnits       int number   inNumber      while  number                 int lastdigit            lastdigit   number   10          number   number 10          inputUnits push back  keyMap lastdigit               if  inputUnits size      2                MakeCombinations inputUnits 0   inputUnits 1   result             else if   inputUnits size    gt  2                 MakeCombinations  inputUnits 0    inputUnits 1   result            vector lt string gt   iterator begin   inputUnits begin                vector lt string gt   iterator end   inputUnits end             begin    2          while   begin    end                         vector lt string gt  intermediate   result              vector lt string gt   iterator ibegin   intermediate begin                 vector lt string gt   iterator iend   intermediate end                  while  ibegin    iend                                MakeCombinations   ibegin    begin  result                     resultbegin                     ibegin                               begin                                    else                   return     int  tmain int argc   TCHAR  argv          keyMap 1            keyMap 2     abc       keyMap 3     def       keyMap 4     ghi       keyMap 5     jkl       keyMap 6     mno       keyMap 7     pqrs       keyMap 8     tuv       keyMap 9     wxyz       keyMap 0             string  inputStr      getline cin  inputStr        int number   0       int length   inputStr length         int tens   1      while  length                 number    tens  inputStr length -1  -  0            length--          tens    10             vector lt string gt  r      ProduceCombinations number  r        cout  lt  lt             vector lt string gt   iterator begin   r begin         vector lt string gt   iterator end   r end         while   begin    end                cout  lt  lt   begin  lt  lt                begin               cout  lt  lt             return 0

User · Answer

This problem is similar to this leetcode problem  Here is the answer I submitted for this problem to leetcode  check github and video for explanation    So the very first thing we need is some way to hold the mappings of a digit and we can use a map for this   private Map lt Integer  String gt  getDigitMap             return Stream of                  new AbstractMap SimpleEntry lt  gt  2   abc                    new AbstractMap SimpleEntry lt  gt  3   def                    new AbstractMap SimpleEntry lt  gt  4   ghi                    new AbstractMap SimpleEntry lt  gt  5   jkl                    new AbstractMap SimpleEntry lt  gt  6   mno                    new AbstractMap SimpleEntry lt  gt  7   pqrs                    new AbstractMap SimpleEntry lt  gt  8   tuv                    new AbstractMap SimpleEntry lt  gt  9   wxyz                    collect Collectors toMap AbstractMap SimpleEntry  getKey                                   AbstractMap SimpleEntry  getValue        The above method is preparing the map and next method I would use is to provide the mapping for provided digit   private String getDigitMappings String strDigit  Map lt Integer String gt  digitMap            int digit   Integer valueOf strDigit           return digitMap containsKey digit    digitMap get digit            This problem can be solved using backtracking and a backtracking solution generally has a structure where the method signature will contain  result container  temp results  original source with index etc  So the method structure would be of the form   private void compute List lt String gt  result  StringBuilder temp  String digits  int start  Map lt Integer  String gt  digitMap              Condition to populate temp value to result           explore other arrangements based on the next input digit           Loop around the mappings of a digit and then to explore invoke the same method recursively           Also need to remove the digit which was in temp at last so as to get proper value in temp for next cycle in loop     And now the method body can be filled as  result will be kept in a list  temp in string builder etc    private void compute List lt String gt  result  StringBuilder temp  String digits  int start  Map lt Integer  String gt  digitMap            if start  gt   digits length         condition             result add temp toString                 return                     String letters   getDigitMappings digits substring start  start   1   digitMap      mappings of a digit to loop around         for  int i   0  i  lt  letters length    i                  temp append letters charAt i                compute result  temp  digits  start 1  digitMap     explore for remaining digits             temp deleteCharAt temp length   - 1      remove last in temp               And finally the method can be invoked as   public List lt String gt  letterCombinations String digits            List lt String gt  result   new ArrayList lt  gt             if digits    null    digits length      0  return result          compute result  new StringBuilder    digits  0  getDigitMap             return result      Now the max mapped chars for a digit can be 4  e g  9 has wxyz  and backtracking involves exhaustive search to explore all possible arrangements  state space tree  so for a digit of size n we are going to have 4x4x4    n times i e  complexity would be O 4 n

User · Answer

namespace WordsFromPhoneNumber            lt summary gt          Summary description for WordsFromPhoneNumber          lt  summary gt       TestClass      public class WordsFromPhoneNumber               private static string   Chars      0    1    ABC    DEF    GHI    JKL    MNO    PQRS    TUV    WXYZ             public WordsFromPhoneNumber                                           TODO  Add constructor logic here                                    region overhead          private TestContext testContextInstance                lt summary gt             Gets or sets the test context which provides            information about and functionality for the current test run              lt  summary gt          public TestContext TestContext                       get                               return testContextInstance                            set                               testContextInstance   value                                    region Additional test attributes                       You can use the following additional attributes as you write your tests                        Use ClassInitialize to run code before running the first test in the class             ClassInitialize               public static void MyClassInitialize TestContext testContext                            Use ClassCleanup to run code after all tests in a class have run             ClassCleanup               public static void MyClassCleanup                             Use TestInitialize to run code before running each test              TestInitialize               public void MyTestInitialize                             Use TestCleanup to run code after each test has run             TestCleanup               public void MyTestCleanup                           endregion           TestMethod          public void TestMethod1                         IList lt string gt  words   Words new int     2                 Assert IsNotNull words   null                Assert IsTrue words Count    3   count                Assert IsTrue words 0      A    a                Assert IsTrue words 1      B    b                Assert IsTrue words 2      C    c                        TestMethod          public void TestMethod23                         IList lt string gt  words   Words new int     2   3                Assert IsNotNull words   null                Assert AreEqual words Count   9   count                Assert AreEqual words 0     AD    AD                Assert AreEqual words 1     AE    AE                Assert AreEqual words 2     AF    AF                Assert AreEqual words 3     BD    BD                Assert AreEqual words 4     BE    BE                Assert AreEqual words 5     BF    BF                Assert AreEqual words 6     CD    CD                Assert AreEqual words 7     CE    CE                Assert AreEqual words 8     CF    CF                        TestMethod          public void TestAll                         int   number   new int  4               Generate number  0                      private void Generate int   number  int index                        for  int x   0  x  lt   9  x    3                                number index    x                  if  index    number Length - 1                                        var w   Words number                       Assert IsNotNull w                       foreach  var xx in number                                                Console Write xx ToString                                               Console WriteLine   possible words  n                        foreach  var ww in w                                                Console Write   0     ww                                             Console WriteLine   n n n                                      else                                       Generate number  index   1                                                       endregion          private IList lt string gt  Words int   number                        List lt string gt  words   new List lt string gt  100               Assert IsNotNull number   null                Assert IsTrue number Length  gt  0   length                StringBuilder word   new StringBuilder number Length               AddWords number  0  word  words                return words                     private void AddWords int   number  int index  StringBuilder word  List lt string gt  words                        Assert IsTrue index  lt  number Length   index  lt  length                Assert IsTrue number index   gt   0   number  gt   0                Assert IsTrue number index   lt   9   number  lt   9                 foreach  var c in Chars number index   ToCharArray                                  word Append c                   if  index  lt  number Length - 1                                        AddWords number  index   1  word  words                                     else                                       words Add word ToString                                       word Length   word Length - 1

User · Answer

In C   recursive    string pattern       0         ABC   DEF   GHI   JKL   MNO   PQRS   TUV   WXYZ    ofstream keyout  keypad txt    void print keypad char  str  int k  vector lt char gt  patt  int i   if str k       0         int x   str k  -  0       for int l   0  l  lt  pattern x  length    l                  patt i    pattern x  l           print keypad str  k 1  patt  i 1             keyout  lt  lt  endl    else if i    k        string st patt data         keyout  lt  lt  st  lt  lt  endl      return        This function can be called with  k  and  i  equal to zero   Anyone  who requires more illustration to understand the logic  can combine recursion technique with following output   ADG ADH ADI  AEG AEH AEI  AFG AFH AFI   BDG BDH BDI  BEG BEH BEI  BFG BFH

User · Answer

In JavaScript  A CustomCounter class takes care of incrementing indexes  Then just output the different possible combinations   var CustomCounter   function min  max        this min   min slice 0      this max   max slice 0      this curr   this min slice 0      this length   this min length    CustomCounter prototype increment   function         for  var i   this length - 1  ii   0  i  gt   ii  i--            this curr i     1         if  this curr i   gt  this max i                 this curr i    0           else               break                    CustomCounter prototype is max   function         for  var i   0  ii   this length  i  lt  ii    i            if  this curr i      this max i                 return false                     return true    var PhoneNumber   function phone number        this phone number   phone number     this combinations         PhoneNumber number to combinations         1    1       2    2    a    b    c       3    3    d    e    f       4    4    g    h    i       5    5    j    k    l       6    6    m    n    o       7    7    p    q    r    s       8    8    t    u    v       9    9    w    x    y    z       0    0           PhoneNumber prototype get combination by digit   function digit        return PhoneNumber number to combinations digit     PhoneNumber prototype add combination by indexes   function indexes        var combination          for  var i   0  ii   indexes length  i  lt  ii    i            var phone number digit   this phone number i          combination    this get combination by digit phone number digit  indexes i              this combinations push combination     PhoneNumber prototype update combinations   function         var min indexes              max indexes           for  var i   0  ii   this phone number length  i  lt  ii    i            var digit   this phone number i          min indexes push 0          max indexes push this get combination by digit digit  length - 1             var c   new CustomCounter min indexes  max indexes       while true            this add combination by indexes c curr          c increment            if  c is max                  this add combination by indexes c curr              break                    var phone number   new PhoneNumber  120   phone number update combinations   console log phone number combinations

User · Answer

This version in C  is reasonably efficient  and it works for non-western digits  like           for example    static void Main string   args        string phoneNumber   null      if  1  lt   args Length          phoneNumber   args 0       if  string IsNullOrEmpty phoneNumber                 Console WriteLine  No phone number supplied             return            else               Console WriteLine  Alphabetic phone numbers for    0       phoneNumber           foreach  string phoneNumberText in GetPhoneNumberCombos phoneNumber               Console Write   0  t   phoneNumberText            public static IEnumerable lt string gt  GetPhoneNumberCombos string phoneNumber        phoneNumber   RemoveNondigits phoneNumber       if  string IsNullOrEmpty phoneNumber           return new List lt string gt          char   combo   new char phoneNumber Length       return GetRemainingPhoneNumberCombos phoneNumber  combo  0      private static string RemoveNondigits string phoneNumber        if  phoneNumber    null          return null      StringBuilder sb   new StringBuilder        foreach  char nextChar in phoneNumber          if  char IsDigit nextChar               sb Append nextChar       return sb ToString       private static IEnumerable lt string gt  GetRemainingPhoneNumberCombos string phoneNumber  char   combo  int nextDigitIndex        if  combo Length - 1    nextDigitIndex                foreach  char nextLetter in phoneNumberAlphaMapping  int char GetNumericValue phoneNumber nextDigitIndex                           combo nextDigitIndex    nextLetter              yield return new string combo                       else               foreach  char nextLetter in phoneNumberAlphaMapping  int char GetNumericValue phoneNumber nextDigitIndex                           combo nextDigitIndex    nextLetter              foreach  string result in GetRemainingPhoneNumberCombos phoneNumber  combo  nextDigitIndex   1                   yield return result                      private static char     phoneNumberAlphaMapping   new char           new char      0         new char      1         new char      a    b    c         new char      d    e    f         new char      g    h    i         new char      j    k    l         new char      m    n    o         new char      p    q    r    s         new char      t    u    v         new char      w    x    y    z

User · Answer

private List lt string gt  strs   new List lt string gt                 char   numbersArray      private int End   0      private int numberOfStrings       private void PrintLetters string numbers                this End   numbers Length          this numbersArray   numbers ToCharArray            this PrintAllCombinations this GetCharacters this numbersArray 0    string Empty  0              private void PrintAllCombinations char   letters  string output  int depth                depth            for  int i   0  i  lt  letters Length  i                          if  depth    this End                                output    letters i                   this PrintAllCombinations this GetCharacters Convert ToChar this numbersArray depth     output  depth                   output   output Substring 0  output Length - 1                             else                               Console WriteLine output   letters i       numberOfStrings                                       private char   GetCharacters char x                char   arr          switch  x                        case  0   arr   new char 1                           return arr              case  1   arr   new char 1                           return arr              case  2   arr   new char 3     a    b    c                     return arr              case  3   arr   new char 3     d    e    f                     return arr              case  4   arr   new char 3     g    h    i                     return arr              case  5   arr   new char 3     j    k    l                     return arr              case  6   arr   new char 3     m    n    o                     return arr              case  7   arr   new char 4     p    q    r    s                     return arr              case  8   arr   new char 3     t    u    v                     return arr              case  9   arr   new char 4     w    x    y    z                     return arr              default  return null

User · Answer

Scala solution   def mnemonics phoneNum  String  dict  IndexedSeq String    Iterable String        def mnemonics d  Int  prefix  String   Seq String          if  d  gt   phoneNum length          Seq prefix        else         for           ch  lt - dict phoneNum charAt d  asDigit          num  lt - mnemonics d   1  s  prefix ch           yield num              mnemonics 0          Assuming each digit maps to at most 4 characters  the number of recursive calls T satisfy the inequality T n   lt   4T n-1   which is of the order 4 n

User · Answer

Here is one for pain C   This one is likley not efficet  in fact I don t think it is at all    But there are some intresting aspects to it    It take a number and converts it into a string It just raises the seed number once each time to create the next sequential string   Whats nice about this is that there is no real limit to the size of the string  no character limit   This allows you to generate longer and longer strings if need be just by waiting    include  lt stdlib h gt   include  lt stdio h gt    define CHARACTER RANGE 95      The range of valid password characters  define CHARACTER OFFSET 32     The offset of the first valid character  void main int argc  char  argv    char  env          int i       long int string      long int worker      char  guess         Current Generation     guess    char  malloc 1         Allocate it so free doesn t fail      int cur       for   string   0    string                   worker   string           free guess           guess    char  malloc  string CHARACTER RANGE 1  sizeof char        Alocate for the number of characters we will need          for   i   0  worker  gt  0  amp  amp  i  lt  string CHARACTER RANGE   1  i            Work out the string                       cur   worker   CHARACTER RANGE      Take off the last digit             worker    worker - cur    CHARACTER RANGE       Advance the digits             cur    CHARACTER OFFSET               guess i    cur                    guess i 1      0        NULL terminate our string          printf   s t d n   guess  string

User · Answer

A Python solution is quite economical  and because it uses generators is efficient in terms of memory use   import itertools  keys   dict enumerate    ABC DEF GHI JKL MNO PQRS TUV WXYZ  split         def words number       digits   map int  str number       for ls in itertools product  map keys get  digits            yield    join ls   for w in words 258       print w   Obviously itertools product is solving most of the problem for you  But writing it oneself is not difficult  Here s a solution in go  which is careful to re-use the array result to generate all solutions in  and a closure f to capture the generated words  Combined  these give O log n  memory use inside product   package main  import        bytes       fmt       strconv     func product choices     byte  result   byte  i int  f func   byte         if i    len result            f result          return           for    c    range choices i            result i    c         product choices  result  i 1  f           var keys   bytes Split   byte    ABC DEF GHI JKL MNO PQRS TUV WXYZ      byte        func words num int  f func   byte         ch        byte       for    b    range strconv Itoa num            ch   append ch  keys b- 0              product ch  make   byte  len ch    0  f     func main         words 256  func b   byte    fmt Println string b

[algorithm] How can I print out all possible letter combinations a given phone number can represent?

Examples related to algorithm

Examples related to language-agnostic

Examples related to combinatorics