Valid Perfect Square - Implementation

Here's the solution:

function isPerfectSquare(num)
    if num = 1 then
        return true
    left := 1
    right := num
    while left ≤ right
        mid := left + (right - left) / 2
        square := mid × mid
        if square = num then
            return true
        else if square < num then
            left := mid + 1
        else
            right := mid - 1
    return false

This runs in $O(\log num)$ time. Binary search halves the search space each iteration, giving logarithmic complexity.