Here are the details of the problem:
Given an array arr of N elements, A majority element in an array arr of size N is an element that appears more than N/2 times in the array. The task is to write a function say isMajority() that takes an array (arr[] ), array’s size (n) and a number to be searched (x) as parameters and returns true if x is a majority element (present more than n/2 times).
The solution states that the lastIndex in the loop should be conditioned.

 static boolean isMajority(int arr[], int n, int x)
    {
        int i, last_index = 0;
 
        /* get last index according to n (even or odd) */
        last_index = (n%2==0)? n/2: n/2+1;
 
        /* search for first occurrence of x in arr[]*/
        for (i = 0; i < last_index; i++)
        {
            /* check if x is present and is present more
               than n/2 times */
            if (arr[i] == x && arr[i+n/2] == x)
                return true;
        }
        return false;
    }

I tried something else and it worked for some examples I have tried. But I am not sure whether it should really work with all cases or not.

public static boolean isMajority(int[]array,int x){
        int lastIndex=array.length-(array.length/2+1);
        for(int i=0;i<=lastIndex;i++){
            if(array[i]==x && array[i]==array[i+array.length/2]){
                return true;
            }
        }
        return false;
    }