just an optimization, even though in this case your code is fine: if you try an array with negatives, then you ll never get inside your if-statements. so the numMaxHeight will never change, so you dont really compute it in such cases. so this solution for an array with negatives, where the max negative appears more than once, will fail. eg [-1, -1]

proposed solution: set numMaxHeight = 0 on initialization

Pass all Testcases..

static int birthdayCakeCandles(int[] ar) {
    Arrays.sort(ar);
    int result = 1;
    int decriment = 2;
    while((decriment <= ar.length) && (ar[ar.length-1] == ar[ar.length-decriment])){
        result++;
        decriment++;
    }
    return result;
}

Try This

static int birthdayCakeCandles(int[] ar) {

    int count =0;
    Arrays.sort(ar);
    int maxHeight = ar[(ar.length)-1];
    for(int i=0;i<ar.length;i++)
    {
        if(ar[i]==maxHeight)
        {
            count++;
        }
    }

    return count;

}

need to add one more check in while condition for all the array elements with same value

while ( i > -1 && a[i--] == tallest ) { count++; }

Count should be initialized to 1 as there is atleast one tallest. In this case, if the frequency of largest element is 0, count would return 0 which is an error

Mine fails on that too. ar.sort(); ar.reverse(); let m = 0; let k = ar[0]; while (parseInt(ar[m]) === parseInt(k)){ m++; } return m; }

while (i>-1&&a[i--] == tallest) { count++; }

You just forgot to check whether the index i is within the bounds.