Introduction

In this blog post, we will explore how to solve the LeetCode problem "633" 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 non-negative integer c, decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: c = 5 Output: true Explanation: 1 1 + 2 2 = 5 Example 2: Input: c = 3 Output: false Constraints: 0 <= c <= 231 - 1

Explanation

Here's the breakdown of the solution:

• Two-Pointer Approach: We use two pointers, left and right, initialized to 0 and the square root of c, respectively.
• Iteration and Adjustment: We iterate while left is less than or equal to right. In each iteration, we calculate the sum of squares of left and right.

• Comparison: If the sum equals c, we return True. If the sum is less than c, we increment left. If the sum is greater than c, we decrement right.

• Runtime Complexity: O(sqrt(c)), Storage Complexity: O(1)

Code

    import math

def judgeSquareSum(c: int) -> bool:
    """
    Given a non-negative integer c, decide whether there're two integers a and b such that a^2 + b^2 = c.
    """
    left = 0
    right = int(math.sqrt(c))

    while left <= right:
        sum_of_squares = left * left + right * right
        if sum_of_squares == c:
            return True
        elif sum_of_squares < c:
            left += 1
        else:
            right -= 1

    return False