Introduction

In this blog post, we will explore how to solve the LeetCode problem "367" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a positive integer num, return true if num is a perfect square or false otherwise. A perfect square is an integer that is the square of an integer. In other words, it is the product of some integer with itself. You must not use any built-in library function, such as sqrt. Example 1: Input: num = 16 Output: true Explanation: We return true because 4 4 = 16 and 4 is an integer. Example 2: Input: num = 14 Output: false Explanation: We return false because 3.742 3.742 = 14 and 3.742 is not an integer. Constraints: 1 <= num <= 231 - 1

Explanation

Here's the solution to determine if a number is a perfect square without using built-in square root functions:

• Binary Search: Employ binary search on the range [1, num] to efficiently find the potential square root.
• Midpoint Check: In each iteration, calculate the square of the midpoint. If the square equals num, return true.

• Adjust Search Range: If the square is less than num, narrow the search to the upper half. Otherwise, narrow it to the lower half. If the binary search completes without finding an exact square, return false.

• Runtime Complexity: O(log n), where n is the input number num. Storage Complexity: O(1).

Code

    def isPerfectSquare(num: int) -> bool:
    """
    Given a positive integer num, return true if num is a perfect square or false otherwise.
    You must not use any built-in library function, such as sqrt.

    Example 1:
    Input: num = 16
    Output: true
    Explanation: We return true because 4 * 4 = 16 and 4 is an integer.

    Example 2:
    Input: num = 14
    Output: false
    Explanation: We return false because 3.742 * 3.742 = 14 and 3.742 is not an integer.

    Constraints:
    1 <= num <= 231 - 1
    """

    left, right = 1, num

    while left <= right:
        mid = left + (right - left) // 2
        square = mid * mid

        if square == num:
            return True
        elif square < num:
            left = mid + 1
        else:
            right = mid - 1

    return False