Find the most popular element in int array

Question

int   a   new int 10  1 2 3 4 5 6 7 7 7 7     how can I write a method and return 7   I want to keep it native without the help of lists  maps or other helpers  Only arrays

User · Answer

You can count the occurrences of the different numbers, then look for the highest one. This is an example that uses a Map, but could relatively easily be adapted to native arrays.

Second largest element: Let us take example : [1,5,4,2,3] in this case, Second largest element will be 4.

Sort the Array in descending order, once the sort done output will be A = [5,4,3,2,1]
Get the Second Largest Element from the sorted array Using Index 1. A[1] -> Which will give the Second largest element 4.

private static int getMostOccuringElement(int[] A) { Map occuringMap = new HashMap();

    //count occurences
    for (int i = 0; i < A.length; i++) { 
        if (occuringMap.get(A[i]) != null) {
            int val = occuringMap.get(A[i]) + 1;
            occuringMap.put(A[i], val);
        } else {
            occuringMap.put(A[i], 1);
        }
    }

    //find maximum occurence
    int max = Integer.MIN_VALUE; 
    int element = -1;
    for (Map.Entry<Integer, Integer> entry : occuringMap.entrySet()) {
        if (entry.getValue() > max) {
            max = entry.getValue();
            element = entry.getKey();
        }
    }
    return element;
}

User · Answer

Take a map to map element  -   count Iterate through array and process the map Iterate through map and find out the popular

User · Answer

import java util Scanner    public class Mostrepeatednumber       public static void main String args                  int most   0          int temp 0          int count 0 tempcount          Scanner in new Scanner System in           System out println  Enter any number            int n in nextInt            int arr   new int n           System out print  Enter array value             for int i 0 i lt  n-1 i                          int n1 in nextInt                arr i  n1                               user input concept closed           logic can be started         for int j 0 j lt  n-1 j                      temp arr j           tempcount 0              for int k 1 k lt  n-1 k                                      if temp  arr k                                                 tempcount                                                     if count lt tempcount                                                                most arr k                                       count tempcount                                                                     System out println most

User · Answer

int largest   0  int k   0  for  int i   0  i  lt  n  i          int count   1      for  int j   i   1  j  lt  n  j              if  a i     a j                 count                        if  count  gt  largest            k   a i           largest   count            So here n is the length of the array  and a   is your array   First  take the first element and check how many times it is repeated and increase the counter  count  as to see how many times it occurs  Check if this maximum number of times that a number has so far occurred if yes  then change the largest variable  to store the maximum number of repetitions  and if you would like to store the variable as well  you can do so in another variable  here k    I know this isn t the fastest  but definitely  the easiest way to understand

User · Answer

This is the wrong syntax  When you create an anonymous array you MUST NOT give its size   When you write the following code        new int    1 23 4 4 5 5 5     You are here creating an anonymous int array whose size will be determined by the number of values that you provide in the curly braces   You can assign this a reference as you have done  but this will be the correct syntax for the same  -      int   a   new int   1 2 3 4 5 6 7 7 7 7     Now  just Sysout with proper index position       System out println a 7

User · Answer

public int getPopularElement int   a      int count   1  tempCount    int popular   a 0     int temp   0    for  int i   0  i  lt   a length - 1   i            temp   a i       tempCount   0      for  int j   1  j  lt  a length  j                if  temp    a j           tempCount              if  tempCount  gt  count              popular   temp        count   tempCount              return popular

User · Answer

public static void main String   args         int   myArray    1 5 4 4 22 4 9 4 4 8       Map lt Integer Integer gt  arrayCounts   new HashMap lt  gt         Integer popularCount    0      Integer popularValue   0       for int i   0  i  lt  myArray length  i              Integer count   arrayCounts get myArray i            if  count    null                count   0                    arrayCounts put myArray i   count    0   1     count           if  count  gt  popularCount                popularCount   count              popularValue   myArray i                        System out println popularValue     -- gt      popularCount

User · Answer

below code can be put inside a main method         TODO Auto-generated method stub     Integer   a     11  1  1  1  1  1  1  1  1  1  2  3  4  1  2  2  2  2  3  4  2        List lt Integer gt  list   new ArrayList lt Integer gt  Arrays asList a        Set lt Integer gt  set   new HashSet lt Integer gt  list       int highestSeq   0      int seq   0      for  int i   set            int tempCount   0          for  int l   list                if  i    l                    tempCount   tempCount   1                            if  tempCount  gt  highestSeq                    highestSeq   tempCount                  seq   i                                      System out println  highest sequence is     seq     repeated for     highestSeq

User · Answer

public class MostFrequentNumber        public MostFrequentNumber                 int frequentNumber List lt Integer gt  list            int popular   0          int holder   0           for Integer number  list                int freq   Collections frequency list number                if holder  lt  freq                   holder   freq                  popular   number                                  return popular              public static void main String   args            int   numbers    4 6 2 5 4 7 6 4 7 7 7            List lt Integer gt  list   new ArrayList lt Integer gt              for Integer num   numbers               list add num                       MostFrequentNumber mostFrequentNumber   new MostFrequentNumber             System out println mostFrequentNumber frequentNumber list

User · Answer

Assuming your array is sorted  like the one you posted  you could simply iterate over the array and count the longest segment of elements  it s something like  narek gevorgyan s post but without the awfully big array  and it uses the same amount of memory regardless of the array s size   private static int getMostPopularElement int   a       int counter   0  curr  maxvalue  maxcounter   -1      maxvalue   curr   a 0        for  int e   a           if  curr    e               counter              else               if  counter  gt  maxcounter                   maxcounter   counter                  maxvalue   curr                            counter   0              curr   e                      if  counter  gt  maxcounter           maxvalue   curr             return maxvalue      public static void main String   args        System out println getMostPopularElement new int   1 2 3 4 5 6 7 7 7 7         If the array is not sorted  sort it with Arrays sort a

User · Answer

public static int getMostCommonElement int   array         Arrays sort array        int frequency   1      int biggestFrequency   1      int mostCommonElement   0       for int i 0  i lt array length-1  i              frequency    array i   array i 1     frequency 1   1          if frequency gt biggestFrequency                biggestFrequency   frequency               mostCommonElement   array i                        return mostCommonElement

User · Answer

Comparing two arrays  I Hope for that this is useful for you   public static void main String   args            int primerArray       1 2 1 3 5           int arrayTow       1 6 7 8            int numberMostRepetly    validateArrays primerArray arrayTow           System out println numberMostRepetly         public static int validateArrays int primerArray    int arrayTow          int numVeces   0       for int i   0  i lt  primerArray length  i              for int c   i 1  c  lt  primerArray length  c                  if primerArray i     primerArray c                    numVeces   primerArray c                      System out println  Numero que mas se repite                      System out println numVeces                                    for int a   0  a  lt  arrayTow length  a                  if numVeces    arrayTow a                      System out println numVeces                   return numVeces                                     return 0

User · Answer

package frequent   import java util HashMap  import java util Map   public class Frequent number          Find the most frequent integer in an array      public static void main String   args            int arr     1 2 3 4 3 2 2 3 3            System out println getFrequent arr            System out println getFrequentBySorting arr                 Using Map   TC  O n   SC  O n      static public int getFrequent int arr             int ans 0          Map lt Integer Integer gt  m   new HashMap lt  gt             for int i arr               if m containsKey i                    m put i  m get i  1                else                  m put i  1                                   int maxVal 0          for Integer in  m keySet                 if m get in  gt maxVal                   ans in                  maxVal   m get in                                   return ans               Sort the array and then find it TC  O nlogn  SC  O 1      public static int getFrequentBySorting int arr             int current arr 0           int ansCount 0          int tempCount 0          int ans current          for int i arr               if i  current                   tempCount                              if tempCount gt ansCount                   ansCount tempCount                  ans i                            current i                    return ans

User · Answer

Best approach will be using map where  key will be element and value will be the count of each element  Along with that keep an array of size that will contain the index of most popular element   Populate this array while map construction itself so that we don t have to iterate through map again   Approach 2 -  If someone want to go with two loop  here is the improvisation from accepted answer where we don t have to start second loop from one every time  public class TestPopularElements       public static int getPopularElement int   a            int count   1  tempCount          int popular   a 0           int temp   0          for  int i   0  i  lt   a length - 1   i                  temp   a i               tempCount   0              for  int j   i 1  j  lt  a length  j                      if  temp    a j                       tempCount                              if  tempCount  gt  count                    popular   temp                  count   tempCount                                  return popular             public static void main String   args            int a     new int    1 2 3 4 5 6 2 7 7 7            System out println  count is    getPopularElement a

User · Answer

This one without maps   public class Main               public static void main String   args            int   a   new int    1  2  3  4  5  6  7  7  7  7            System out println getMostPopularElement a                       private static int getMostPopularElement int   a                         int maxElementIndex   getArrayMaximumElementIndex a            int   b   new int a maxElementIndex    1           for  int i   0  i  lt  a length  i                    b a i                       return getArrayMaximumElementIndex b              private static int getArrayMaximumElementIndex int   a            int maxElementIndex   0           for  int i   1  i  lt  a length  i                  if  a i   gt   a maxElementIndex                     maxElementIndex   i                                   return maxElementIndex                   You only have to change some code if your array can have elements which are  lt  0  And this algorithm is useful when your array items are not big numbers

User · Answer

If you don t want to use a map  then just follow these steps    Sort the array  using Arrays sort    Use a variable to hold the most popular element  mostPopular   a variable to hold its number of occurrences in the array  mostPopularCount   and a variable to hold the number of occurrences of the current number in the iteration  currentCount  Iterate through the array  If the current element is the same as mostPopular  increment currentCount  If not  reset currentCount to 1  If currentCount is   mostPopularCount  set mostPopularCount to currentCount  and mostPopular to the current element

User · Answer

I hope this helps   public class Ideone       public static void main String   args  throws java lang Exception        int   a    1 2 3 4 5 6 7 7 7       int len   a length       System out println len         for  int i   0  i  lt   len - 1  i               while  a i     a i   1                 System out println a i                 break

User · Answer

Using Java 8 Streams   int data       1  5  7  4  6  2  0  1  3  2  2    Map lt Integer  Long gt  count   Arrays stream data       boxed        collect Collectors groupingBy Function identity    counting       int max   count entrySet   stream        max  first  second  - gt            return  int   first getValue   - second getValue                 get   getKey     System out println max     Explanation  We convert the int   data array to boxed Integer Stream  Then we collect by groupingBy on the element and use a secondary counting collector for counting after the groupBy   Finally we sort the map of element - gt  count based on count again by using a stream and lambda comparator

User · Answer

Try this answer  First  the data  int   a    1 2 3 4 5 6 7 7 7 7    Here  we build a map counting the number of times each number appears  Map lt Integer  Integer gt  map   new HashMap lt Integer  Integer gt     for  int i   a        Integer count   map get i       map put i  count    null   count 1   1      Now  we find the number with the maximum frequency and return it  Integer popular   Collections max map entrySet        new Comparator lt Map Entry lt Integer  Integer gt  gt           Override     public int compare Entry lt Integer  Integer gt  o1  Entry lt Integer  Integer gt  o2            return o1 getValue   compareTo o2 getValue              getKey     As you can see  the most popular number is seven  System out println popular    gt  7  EDIT Here s my answer without using maps  lists  etc  and using only arrays  although I m sorting the array in-place  It s O n log n  complexity  better than the O n 2  accepted solution  public int findPopular int   a         if  a    null    a length    0          return 0       Arrays sort a        int previous   a 0       int popular   a 0       int count   1      int maxCount   1       for  int i   1  i  lt  a length  i              if  a i     previous              count            else               if  count  gt  maxCount                    popular   a i-1                   maxCount   count                            previous   a i               count   1                       return count  gt  maxCount   a a length-1    popular

User · Answer

Mine Linear O N   Using map to save all the differents elements found in the array and saving the number of times ocurred  then just getting the max from the map   import java util HashMap  import java util Map  import java util Comparator  import java util HashMap  import java util Map  import java util stream IntStream   public class MosftOftenNumber           for O N    map O 1    O N       public static int mostOftenNumber int   a                final Map m   new HashMap lt Integer Integer gt             int max   0          int element   0           for  int i 0  i lt a length  i                   initializing value for the map the value will have the counter of each element               first time one new number its found will be initialize with zero              if  m get a i      null                  m put a i  0                  save each value from the array and increment the count each time its found             m put a i     Integer  m get a i     1                  check the value from each element and comparing with max             if    Integer  m get a i    gt  max                   max    Integer  m get a i                    element   a i                                    System out println  Times repeated      max           return element             public static int mostOftenNumberWithLambdas int   a                Integer max   IntStream of a  boxed   max Integer  compareTo  get            Integer coumtMax   Math toIntExact IntStream of a  boxed   filter number - gt  number equals max   count             System out println  Times repeated      coumtMax           return max             public static void main String args              int   array    1 1 2 1 1           int   array    2 2 1 2 2           int   array    1 2 3 4 5 6 7 7 7 7           System out println  Most often number with loops      mostOftenNumber array            System out println  Most often number with lambdas      mostOftenNumberWithLambdas array

User · Answer

import java util HashMap  import java util Map  import java lang Integer  import java util Iterator  public class FindMood       public static void main String    args       int arrayToCheckFrom       1 2 4 4 5 5 5 3 3 3 3 3 3 3 3       Map map   new HashMap lt Integer  Integer gt         for int i   0   i  lt  arrayToCheckFrom length  i         int sum   0        for int k   0   k  lt  arrayToCheckFrom length   k               if arrayToCheckFrom i   arrayToCheckFrom k             sum    1                 map put arrayToCheckFrom i   sum             System out println getMaxValue map        public static Integer getMaxValue  Map lt Integer Integer gt  map           Map Entry lt Integer Integer gt  maxEntry   null          Iterator iterator   map entrySet   iterator              while iterator hasNext                 Map Entry lt Integer Integer gt  pair    Map Entry lt Integer Integer gt   iterator next                if maxEntry    null    pair getValue   compareTo maxEntry getValue    gt 0                   maxEntry   pair                                    return maxEntry getKey

User · Answer

Array elements value should be less than the array length for this one   public void findCounts int   arr  int n        int i   0       while  i  lt  n            if  arr i   lt   0                i                continue                     int elementIndex   arr i  - 1           if  arr elementIndex   gt  0                arr i    arr elementIndex               arr elementIndex    -1                    else               arr elementIndex --              arr i    0              i                         Console WriteLine  Below are counts of all elements         for  int j   0  j  lt  n  j              Console WriteLine j   1    - gt     Math Abs arr j               Time complexity of this will be O N  and space complexity will be O 1

User · Answer

Seems like you are looking for the Mode value  Statistical Mode    have a look at Apache s Docs for Statistical functions

User · Answer

public class MostFrequentIntegerInAnArray        public static void main String   args            int   items   new int   2 1 43 1 6 73 5 4 65 1 3 6 1 1           System out println  Most common item     getMostFrequentInt items                 Time Complexity   O N        Space Complexity   O N      public static int getMostFrequentInt int   items           Map lt Integer  Integer gt  itemsMap   new HashMap lt Integer  Integer gt  items length           for int item   items               if  itemsMap containsKey item                   itemsMap put item  1               else                 itemsMap put item  itemsMap get item  1                      int maxCount   Integer MIN VALUE          for Entry lt Integer  Integer gt  entry   itemsMap entrySet                 if entry getValue    gt  maxCount                  maxCount   entry getValue                      return maxCount

User · Answer

Assuming your int array is sorted  i would do       int count   0  occur   0  high   0  a   for  a   1  a  lt  n length  a          if  n a - 1     n a            count           if  count  gt  occur               occur   count             high   n a                   else           count   0           System out println  highest occurence       high

[java] Find the most popular element in int[] array

Examples related to java

Examples related to arrays