# Solving Leetcode Interviews in Seconds with AI: Sum of Square Numbers


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```