Find the element with highest occurrences in an array [java]

Question

I have to find the element with highest occurrences in a double array. I did it like this:

int max = 0;
for (int i = 0; i < array.length; i++) {
       in         


        

Answer 1

Here's a java solution --

        List<Integer> list = Arrays.asList(1, 2, 2, 3, 2, 1, 3);
        Set<Integer> set = new HashSet(list);
        int max = 0;
        int maxtemp;
        int currentNum = 0;
        for (Integer k : set) {
            maxtemp = Math.max(Collections.frequency(list, k), max);
            currentNum = maxtemp != max ? k : currentNum;
            max = maxtemp;
        }
        System.out.println("Number :: " + currentNum + " Occurs :: " + max + " times");

Answer 2

You can solve this problem in one loop with without using HashMap or any other data structure in O(1) space complexity.

Initialize two variables count = 0 and max = 0 (or Integer.MIN_VALUE if you have negative numbers in your array)

The idea is you will scan through the array and check the current number,

• if it is less than your current max...then do nothing
• if it is equal to your max ...then increment the count variable
• if it is greater than your max..then update max to current number and set count to 1

Code:

int max = 0, count = 0;
for (int i = 0; i < array.length; i++) {
    int num = array[i];
    if (num == max) {
        count++;
    } else if (num > max) {
        max = num;
        count = 1;
    }
}

Answer 3

Use Collections.frequency option:

 List<String> list = Arrays.asList("1", "1","1","1","1","1","5","5","12","12","12","12","12","12","12","12","12","12","8");
      int max = 0;
      int curr = 0;
      String currKey =  null;
      Set<String> unique = new HashSet<String>(list);

          for (String key : unique) {
                curr = Collections.frequency(list, key);

               if(max < curr){
                 max = curr;
                 currKey = key;
                }
            }

          System.out.println("The number "  + currKey + " happens " + max + " times");

Output:

The number 12 happens 10 times

Answer 4

This is how i have implemented in java..

import java.io.*;
class Prog8
{
    public static void main(String[] args) throws IOException 
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        System.out.println("Input Array Size:");
        int size=Integer.parseInt(br.readLine());
        int[] arr= new int[size];
        System.out.println("Input Elements in Array:");
        for(int i=0;i<size;i++)
            arr[i]=Integer.parseInt(br.readLine());
        int max = 0,pos=0,count = 0;
        for (int i = 0; i < arr.length; i++)
        {
            count=0;
            for (int j = 0; j < arr.length; j++) 
            {
                if (arr[i]==arr[j])
                    count++;
            }
            if (count >=max)
            {
                max = count;
                pos=i;
            }
        }

        if(max==1)
            System.out.println("No Duplicate Element.");
        else
            System.out.println("Element:"+arr[pos]+" Occourance:"+max);
    }
}

Answer 5

The solution with Java 8

       int result = Arrays.stream(array)
           .boxed()
           .collect(Collectors.groupingBy(i->i,Collectors.counting()))
           .values()
           .stream()
           .max(Comparator.comparingLong(i->i))
           .orElseThrow(RuntimeException::new));

Answer 6

Here is Ruby SOlution:

def maxOccurence(arr)
  m_hash = arr.group_by(&:itself).transform_values(&:count)
  elem = 0, elem_count = 0
  m_hash.each do |k, v|
    if v > elem_count
        elem = k
        elem_count = v
    end
  end
  "#{elem} occured #{elem_count} times"
end

p maxOccurence(["1", "1","1","1","1","1","5","5","12","12","12","12","12","12","12","12","12","12","8"])

output:

"12 occured 10 times"