Palindrome check in Javascript

Question

I have the following   function checkPalindrom palindrom         for  var i   palindrom length  i  gt  0  i--                 if  palindrom i    palindrom charAt palindrom length -1                         document write  the word is palindrome              else              document write  the word is not palindrome                       checkPalindrom  wordthatwillbechecked      What is wrong with my code  I want to check if the word is a palindrome

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SHORTEST CODE  31 chars  ES6    p s  gt s      s  reverse   join   p  racecar      true   Keep in mind short code isn t necessarily the best  Readability and efficiency can matter more

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function palindromCheck str        var palinArr  i          palindrom            palinArr   str split    s        -    ig        for  i   0  i  lt  palinArr length  i              if  palinArr i  toLowerCase       palinArr i  split     reverse   join     toLowerCase    amp  amp              palinArr i                        palindrom push palinArr i                            return palindrom join          console log palindromCheck  There is a man  his name  was Bob        a  Bob   Finds and upper to lower case  Split string into array  I don t know why a few white spaces remain  but I wanted to catch and single letters

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in palindrom i    palindrom charAt palindrom length -1 should be    or     palindrom charAt palindrom length -1 should be palindrom charAt palindrom length - i

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Writing the code for checking palindromes following the best practices of JavaScript    x000D   x000D   function    x000D    use strict   x000D    x000D   var testStringOne    Madam   x000D   var testStringTwo    testing   x000D   var testNull   null  x000D   var testUndefined  x000D    x000D   console log checkIfPalindrome testStringOne    x000D   console log checkIfPalindrome testStringTwo    x000D   console log checkIfPalindrome testNull    x000D   console log checkIfPalindrome testUndefined    x000D    x000D   function checkIfPalindrome testStringOne   x000D     x000D    if  testStringOne   x000D     return false  x000D      x000D     x000D    var testArrayOne   testStringOne toLowerCase   split      x000D    testArrayOne reverse    x000D     x000D    if testArrayOne toString      testStringOne toLowerCase   split     toString     x000D     return true  x000D     else  x000D     return false  x000D      x000D     x000D    x000D        x000D   x000D   x000D

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I used a ternary in this case to keep track of the index backwards   function palindromeCheck string      let str   string toLowerCase      let palindrome     for  let i   0  i  lt  str length  i          if  str i      str str length -  i    0   1   i   1          palindrome   true        else         palindrome   false

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The logic here is not quite correct  you need to check every letter to determine if the word is a palindrome  Currently  you print multiple times  What about doing something like   function checkPalindrome word            var l   word length      for  var i   0  i  lt  l   2  i              if  word charAt i      word charAt l - 1 - i                 return false                      return true     if  checkPalindrome  1122332211          document write  The word is a palindrome      else       document write  The word is NOT a palindrome        Which should print that it IS indeed a palindrome

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First problem    is assign    is compare  Second problem  Your logic here is wrong  palindrom charAt palindrom length -1   You are subtracting one from the charAt and not the length   Third problem  it still will be wrong since you are not reducing the length by i

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Palindrome is a string  which when read in both   forward and backward way is same  Example  Example  madam  lol  pop    radar etc    Simple Logic    Sample string   let str1    radar     breaking into a char array and sort   let str1Arr   str1 split     sort     a a d r r   let str2Arr   str1 split     reverse   sort     a a d r r   now join both string separately and compare  if both are same then it is pelindrome        if str1Arr join        str2Arr join                 console log  Palindrome           else         console log  Not Palindrome

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I thought I d share my own solution    x000D   x000D  function palindrome string   x000D      var reverseString       x000D      for var k in string   x000D         reverseString    string  string length - k  - 1   x000D        x000D    if string     reverseString   x000D      console log  Hey there palindrome    x000D     else  x000D      console log  You are not a palindrome    x000D      x000D    x000D  palindrome  ana    x000D   x000D   x000D    Hope will help someone

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Some above short anwsers is good  but it s not easy for understand  I suggest one more way   function checkPalindrome inputString         if inputString length    1           return true       else          var i   0          var j   inputString length -1          while i  lt  j               if inputString i     inputString j                    return false                            i                j--                      return true      I compare each character  i start form left  j start from right  until their index is not valid  i lt j    It s also working in any languages

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It works to me  function palindrome str         remove special characters  spaces and make lowercase     var removeChar   str replace    A-Z0-9  ig      toLowerCase          reverse removeChar for comparison     var checkPalindrome   removeChar split     reverse   join            Check to see if str is a Palindrome      return  removeChar     checkPalindrome

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I ve added some more to the above functions  to check strings like   Go hang a salami  I m a lasagna hog     function checkPalindrom str        var str   str replace    a-zA-Z0-9   gi      toLowerCase        return str    str split     reverse   join          Thanks

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I m wondering why nobody suggested this    ES6       aba  - gt  true     acb  - gt  false     aa  - gt  true     abba  - gt  true     s  - gt  true isPalindrom    str         gt      if  str 0      str str length - 1         return str length  lt   1   true   isPalindrom str slice 1  -1          return false     alert   aba    acb    aa    abba    s   map  e  i    gt  isPalindrom e   join      ES5       aba  - gt  true     acb  - gt  false     aa  - gt  true     abba  - gt  true     s  - gt  true function isPalindrom str    gt      var str   typeof str      string         str     if  str 0      str str length - 1         return str length  lt   1   true   isPalindrom str slice 1  -1          return false     alert   aba    acb    aa    abba    s   map function  e  i        return isPalindrom e      join

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function checkPalindrom palindrom       var flag   true     var j   0      for  var i   palindrom length-1  i  gt  palindrom length   2  i--                 if  palindrom i     palindrom j                         flag   false             break     why this  It ll exit the loop at once when there is a mismatch                    j            if  flag       document write  the word is palindrome           else   document write  the word is not palindrome           checkPalindrom  wordthatwillbechecked      Why am I printing the result outside the loop  Otherwise  for each match in the word  it ll print  is or is not pallindrome  rather than checking the whole word   EDIT  Updated with changes and a fix suggested by Basemm

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function checkPalindrome  str        var str   str toLowerCase        var original   str split      join          var reversed   original split      reverse   join         return  original     reversed

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You can try the following  function checkPalindrom  str          str   str toLowerCase          return str    str split     reverse   join                 if checkPalindrom  Racecar              console log  Palindrome          else           console log  Not Palindrome

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I found this on an interview site       Write an efficient function that checks whether any permutation of an   input string is a palindrome  You can ignore punctuation  we only care   about the characters    Playing around with it I came up with this ugly piece of code     function checkIfPalindrome text        var found           var foundOne   0      text   text replace    a-z0-9  gi      toLowerCase        for  var i   0  i  lt  text length  i              if  found text i                  found text i                else               found text i     1                      for  var x in found            if  found x      1                foundOne                if  foundOne  gt  1                    return false                                    for  var x in found            if  found x   gt  2  amp  amp  found x    2  amp  amp  foundOne                return false                      return true      Just leaving it here for posterity

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As a much clearer recursive function  http   jsfiddle net dmz2x117   function isPalindrome letters         var characters    letters split              firstLetter   characters shift            lastLetter    characters pop         if  firstLetter     lastLetter            return false             if  characters length  lt  2            return true             return isPalindrome characters join

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How about this  using a simple flag   function checkPalindrom str      var flag   true     for  var i   0  i  lt   str length-1  i         if  str i      str str length - i-1          flag   false                     if flag    false         console log  the word is not a palindrome                  else      console log  the word is a palindrome              checkPalindrom  abcdcba

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Maybe I will suggest alternative solution   function checkPalindrom  str      return str    str split     reverse   join          UPD  Keep in mind however that this is pretty much  cheating  approach  a demonstration of smart usage of language features  but not the most practical algorithm  time O n   space O n    For real life application or coding interview you should definitely use loop solution  The one posted by Jason Sebring in this thread is both simple and efficient  time O n   space O 1

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Sharing my fast variant which also support spaces   x000D   x000D  function isPalindrom str    x000D    var ia   0  x000D    var ib   str length - 1  x000D    do   x000D      if  str ia      str ib   continue  x000D   x000D         if spaces skip  amp  retry x000D      if  str ia           amp  amp  ib    continue  x000D      if  str ib           amp  amp  ia--  continue  x000D   x000D      return false  x000D      while    ia  lt  --ib   x000D    return true  x000D    x000D  var palindrom  never odd or even   x000D  var res   isPalindrom palindrom   x000D  document getElementById  check   innerHTML       palindrom         checked to be     res  x000D   lt span id  check    gt  x000D   x000D   x000D

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lt script gt      function checkForPally             var input   document getElementById  inputTable   value          var input   input replace   s g                var arrayInput   input split             var inputReversed   arrayInput reverse            var stringInputReversed   inputReversed join               if  input    stringInputReversed                check value    The word you enter is a palindrome                    if  input    stringInputReversed                check value    The word you entered is not a palindrome                   lt  script gt    You first use the getElement tag to set the initial variable of the palindrome  Seeing as a palindrome can be multiple words you use the regex entry to remove the whitespaces  You then use the split function to convert the string into an array   Next up  you use the reverse method to reverse the array when this is done you join the reversed array back into a string  Lastly  you just use a very basic if statement to check for equality  if the reversed value is equall to the initial variable you have yourself a palindrome

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The most important thing to do when solving a Technical Test is Don t use shortcut methods -- they want to see how you think algorithmically  Not your use of methods   Here is one that I came up with  45 minutes after I blew the test   There are a couple optimizations to make though  When writing any algorithm  its best to assume false and alter the logic if its looking to be true   isPalindrome     Basically  to make this run in O N   linear  complexity you want to have 2 iterators whose vectors point towards each other  Meaning  one iterator that starts at the beginning and one that starts at the end  each traveling inward  You could have the iterators traverse the whole array and use a condition to break return once they meet in the middle  but it may save some work to only give each iterator a half-length by default   for loops seem to force the use of more checks  so I used while loops - which I m less comfortable with   Here s the code          TODO  If func counts out  let it return 0       Assume  isPalindrome  invert logic      function isPalindrome S       var s   S         len   s length         mid   len 2          i   0  j   len-1       while i lt mid           var l   s charAt i           while j gt  mid               var r   s charAt j               if l     r                   console log   while     i  l         j  r                   --j                  break                            console log   while     i  l         j  r               return 0                      i            return 1     var nooe   solution  neveroddoreven        even char length var kayak   solution  kayak        odd char length var kayaks   solution  kayaks     console log   isPalindrome   nooe  kayak  kayaks     Notice that if the loops count out  it returns true  All the logic should be inverted so that it by default returns false  I also used one short cut method String prototype charAt n   but I felt OK with this as every language natively supports this method

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This worked for me   var number   8008 number   number       numberreverse   number split     reverse   join      console log   The number if reversed is     numberreverse   if  number    numberreverse      console log  Yes  this is a palindrome    else     console log  Nope  It isnt a palindrome

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recursion only index compare  const isPalindrome    str  start   0  end   str length - 1    gt        if  start  gt   end            return true             if  str start      str end             return isPalindrome str  start   1  end - 1              return false

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Using recursion    function isPalindromeRecursive str      const isLessThan2   str length  lt  2    const firstAndLastEqual   str slice 0  1      str slice -1     return  isLessThan2  amp  amp  firstAndLastEqual        isPalindromeRecursive str slice 1  -1          isLessThan2

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At least three things    You are trying to test for equality with    which is used for setting   You need to test with    or        Probably the latter  if you don t have a reason for the former   You are reporting results after checking each character   But you don t know the results until you ve checked enough characters  You double-check each character-pair  as you really only need to check if  say first     last and not also if last     first

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This avoids regex while also dealing with strings that have spaces and uppercase     function isPalindrome str        str   str split           var str2   str filter function x            if x          amp  amp  x                        return x                         return console log str2 join     toLowerCase       console log str2 reverse   join     toLowerCase         isPalindrome  A car  a man  a maraca      true

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Here is a solution that works even if the string contains non-alphanumeric characters   function isPalindrome str        str   str toLowerCase   replace   W    g           return str    str split     reverse   join

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JavaScript  Using regexp  this checks for alphanumeric palindrome and disregards space and punctuation    function palindrome str      str   str match   A-Za-z0-9  gi  join     toLowerCase             A-Za-z0-9  gi  above makes str alphanumeric    for var i   0  i  lt  Math floor str length 2   i        only need to run for half the string length      if str charAt i      str charAt str length-i-1        uses     to compare characters one-by-one from the beginning and end       return  Try again                return  Palindrome      palindrome  A man  a plan  a canal  Panama       palindrome  4 2     -     2-4       This solution would also work on something like this

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If you want efficiency and simplicity  I recommend this approach   function Palindrome str        let len   str length           i   0       if  len  lt  3           return false       while  len--            if  str i      str len               i            else             return false             return true     console log Palindrome  aba     true console log Palindrome  abc     false

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One more solution with ES6   isPalin   str   gt      str  every  c  i    gt  c     str str length-1-i

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25x faster than the standard answer  function isPalindrome s i     return  i i  0  lt 0  i gt  s length gt  gt 1  s i   s s length-1-i  amp  amp isPalindrome s   i       use like   isPalindrome  racecar      as it defines  i  itself  Fiddle  http   jsfiddle net namcx0yf 9   This is  25 times faster than the standard answer below   function checkPalindrome str      return str    str split     reverse   join          Fiddle  http   jsfiddle net t0zfjfab 2   View console for performance results   Although the solution is difficult to read and maintain  I would recommend understanding it to demonstrate non-branching with recursion and bit shifting to impress your next interviewer   explained  The    and  amp  amp  are used for control flow like  if   else   If something left of    is true  it just exits with true  If something is false left of    it must continue  If something left of  amp  amp  is false  it exits as false  if something left of a  amp  amp  is true  it must continue  This is considered  non-branching  as it does not need if-else interupts  rather its just evaluated   1  Used an initializer not requiring  i  to be defined as an argument  Assigns  i  to itself if defined  otherwise initialize to 0  Always is false so next OR condition is always evaluated    i   i    0   lt  0   2  Checks if  i  went half way but skips checking middle odd char  Bit shifted here is like division by 2 but to lowest even neighbor division by 2 result  If true then assumes palindrome since its already done  If false evaluates next OR condition   i  gt   s length  gt  gt  1   3  Compares from beginning char and end char according to  i  eventually to meet as neighbors or neighbor to middle char  If false exits and assumes NOT palindrome  If true continues on to next AND condition   s i     s s length-1-i    4  Calls itself again for recursion passing the original string as  s   Since  i  is defined for sure at this point  it is pre-incremented to continue checking the string s position  Returns boolean value indicating if palindrome   isPalindrome s   i    BUT     A simple for loop is still about twice as fast as my fancy answer  aka KISS principle   function fastestIsPalindrome str      var len   Math floor str length   2     for  var i   0  i  lt  len  i        if  str i      str str length - i - 1         return false    return true      http   jsfiddle net 6L953awz 1

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function checkPalindrom palindrom      palindrom  palindrom toLowerCase       var flag   true     var j     j    palindrom length  -1        console log j      var cnt   j   2       console log cnt       for  i   0  i  lt  cnt 1   i   j--                 console log  J is   gt    j           console log palindrom i      lt    gt     palindrom j            if  palindrom i     palindrom j                         flag   false             break                       if  flag       console log  the word is palindrome           else   console log  the word is not palindrome           checkPalindrom  Avid diva

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Nice answers here  Here is another approach       function checkPalindrom palindrom         var len   palindrom length    get length of the word       var pos   len-1         get index of the last character       var median    len 2     get median character       if len  lt   1            document write  The word is a Palindrome           else           for var i   0  i  lt  median 1  i                if palindrom charAt i     palindrom charAt pos-i                 document write  The word is a Palindrome                                    document write  The word is not a Palindrome                         checkPalindrom  wordthatwillbechecked

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function palindrome str       Good luck    convert the string into lowerCase   str   str toLowerCase      this line remove all non alphanumeric characters   strAlphaNum   str replace    a-z0-9  g        this line create an array of the strAlphaNum string   strArray   strAlphaNum split        this line reverse the array  strReversed   strArray reverse      this line join the reversed array to make a string whithout space   strRevJoin   strReversed join        this condition check if the given string is a palindrome   if  strAlphaNum     strRevJoin    return true        else  return false

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Recursive Method      var low  var high  var A    abcdcba      function palindrome A   low  high     A   A split        if  low  gt  high      low    high       return true       if A low      A high       A   A join         low   low   1      high   high - 1      return palindrome A   low  high       else     return  not a palindrome         palindrome A  0  A length-1

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Loop through the string characters both forwards  i  and backwards  j  using a for loop  If at any point the character at str i  does not equal str j  - then it is not a palindrome  If we successfully loop through the string then it is a palindrome   function isPalindrome str      for var i   0  j   str length - 1  i  lt  str length  i    j--        if  str i      str j   return false        return true

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This version allows special characters like      or              that in other answers doesn t resolve it   function palindromeUnicode s         var sc   decodeURIComponent escape s         return sc     Array from sc  reverse   join       console log palindromeUnicode      italavalati           true

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Thought I will share my solution using Array prototype filter    filter   filters the array based on boolean values the function returns   var inputArray      a   ab   aba   abab   ababa   var outputArray inputArray filter function isPalindrome x     if  x length lt 2  return true    var y x split     reverse   join        return x  y     console log outputArray

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function myPolidrome polidrome    var string polidrome split     join        for var i 0 i lt string length i         if string length  1        console log  is polidrome        else if string i   string charAt string length-1         console log  is not polidrome         break     else       return  myPolidrome polidrome substring 1 polidrome length-1                myPolidrome  asasdsdsa

[javascript] Palindrome check in Javascript

SHORTEST CODE (31 chars)(ES6):

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