Given an array, How to find the first element that appears an even number of times?

Question

Given an unsorted array which has a number in the majority (a number appears more than 50% in the array), how to find that number?

Answer 1

#include<stdio.h>
void even(int arr[],int n)
{
    int i,count=0,j;
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(arr[i]==arr[j])
            {
                count++;    
            }
        }

        if(count%2==0)
        {
            printf("%d",arr[i]);
            break;
        }
        count=0;        
    }
}

int main()
{
    int arr[50];
    int n,i;
    printf("ENTER THE SIZE OF ARRAY: ");
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        printf("arr[%d]=",i);
        scanf("%d",&arr[i]);

    }
    even(arr,n);
}

Answer 2

Algorithm :
1: scan the array from left to right.
1.a: Make count array to each element of given array.
2. scan count array from left to right.
2.a check current count array element is divisible by 2
2.b Print that element and stop
3: exit.