Valid Perfect Square

Problem

Return true iff num is a perfect square. Do NOT use Math.sqrt.

Constraints

• 1 ≤ num ≤ 2³¹−1

Approach — Binary Search

This is a Binary Search problem. The idea: repeatedly halve the search space, discarding the half that can't contain the answer. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(log n) time.

Solution code

Python

class Solution:
    def isPerfectSquare(self, num):
        lo, hi = 1, num
        while lo <= hi:
            mid = (lo + hi) // 2
            sq = mid * mid
            if sq == num:
                return True
            if sq < num:
                lo = mid + 1
            else:
                hi = mid - 1
        return False

Java

class Solution {
    public boolean isPerfectSquare(int num) {
        long lo = 1, hi = num;
        while (lo <= hi) {
            long mid = lo + (hi - lo) / 2;
            long sq = mid * mid;
            if (sq == num) return true;
            if (sq < num) lo = mid + 1;
            else          hi = mid - 1;
        }
        return false;
    }
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.