Square root of a number

1 <= A <= 10^8

First and the only argument is an integer A.

Return an integer which is the square root of A if A is perfect square otherwise return -1.

Input 1:
A = 4

Input 2:
A = 1001

Output 1:
2

Output 2:
-1

Explanation 1:
sqrt(4) = 2

Explanation 2:
1001 is not a perfect square.

public class Solution {
    public int sqrt(int A) {
        long start = 1, end = A, ans = -1;
        while (start <= end) {
            long mid = (start + end) / 2;
            if (mid * mid == A) return (int)mid;
            if (mid * mid < A) {
                start = mid + 1;
                ans = mid;
            } else {
                end = mid - 1;
            }
        }
        return (int)ans;
    }
}

class Solution:
    def sqrt(self, A):
        start, end, ans = 1, A, -1
        while start <= end:
            mid = (start + end) // 2
            if mid * mid == A:
                return mid
            if mid * mid < A:
                start = mid + 1
                ans = mid
            else:
                end = mid - 1
        return ans

function sqrt(A) {
    let start = 1, end = A, ans = -1;
    while (start <= end) {
        let mid = Math.floor((start + end) / 2);
        if (mid * mid === A) return mid;
        if (mid * mid < A) {
            start = mid + 1;
            ans = mid;
        } else {
            end = mid - 1;
        }
    }
    return ans;
}