# Solving Leetcode Interviews in Seconds with AI: Count Square Sum Triples


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1925" 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
	> A square triple (a,b,c) is a triple where a, b, and c are integers and a2 + b2 = c2. Given an integer n, return the number of square triples such that 1 <= a, b, c <= n.   Example 1:  Input: n = 5 Output: 2 Explanation: The square triples are (3,4,5) and (4,3,5).  Example 2:  Input: n = 10 Output: 4 Explanation: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).    Constraints:  1 <= n <= 250  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Iterate and Check:** Loop through all possible combinations of `a`, `b`, and `c` within the range 1 to `n`.
*   **Square Triple Condition:** For each combination, check if `a*a + b*b == c*c`.
*   **Count Valid Triples:** Increment a counter whenever the condition is met.

*   **Runtime Complexity:** O(n^3). **Storage Complexity:** O(1).

	
	# Code
	```python
	def count_triples(n: int) -> int:
    """
    Given an integer n, return the number of square triples such that 1 <= a, b, c <= n.
    A square triple (a,b,c) is a triple where a, b, and c are integers and a2 + b2 = c2.

    For example:
    count_triples(5) == 2
    count_triples(10) == 4
    count_triples(20) == 8
    """
    count = 0
    for a in range(1, n + 1):
        for b in range(1, n + 1):
            for c in range(1, n + 1):
                if a * a + b * b == c * c:
                    count += 1
    return count
	```