This approach can check perfect square for any positive integer in optimal time i.e, O(log n)

ValidPerfectSquare_v4.java

// this method will return 1 if N is valid perfect square number otherwise -1
public static int validPerfectSquare(long N) {
    if (N == 1) return 1;
    
    long lowPointer = 2, highPointer = N / 2;
    while (lowPointer <= highPointer) {
        long mid = lowPointer + (highPointer - lowPointer) / 2;
        if (mid * mid == N) return 1;
        else if (mid * mid > N) highPointer = mid - 1;
        else lowPointer = mid + 1;
    }
    return -1;
}