367. Valid Perfect Square

// Approach: Binary search for an integer r in [1, num] such that r*r == num.
// Time: O(log n) Space: O(1)

public class Solution
{
    public bool IsPerfectSquare(int num)
    {
        int l = 1;
        int r = num;

        while (l < r)
        {
            int m = (l + r) / 2;
            if (m >= num / m)
                r = m;
            else
                l = m + 1;
        }

        return l * l == num;
    }
}