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Count Square Sum Triples

Difficulty: Easy

Problem Description

Given an integer n, return the number of square triples (a, b, c) such that a² + b² = c² and 1 <= a, b, c <= n.

Key Insights

  • A square triple is defined by the Pythagorean theorem: a² + b² = c².
  • The values for a, b, and c must be constrained between 1 and n.
  • Each valid combination of (a, b) that satisfies a² + b² = c² must result in c being a whole number and within the defined range.

Space and Time Complexity

Time Complexity: O(n²)
Space Complexity: O(1)

Solution

To solve this problem, we use a brute-force approach where we iterate through all possible values of a and b from 1 to n. For each pair (a, b), we calculate c using the formula c = sqrt(a² + b²). We then check if c is an integer and within the range of 1 to n. If it is, we count the valid square triple. This method effectively enumerates all possible pairs and checks the necessary condition.

Code Solutions

def countSquareSumTriples(n):
    count = 0
    for a in range(1, n + 1):
        for b in range(1, n + 1):
            c = (a**2 + b**2) ** 0.5  # Calculate c
            if c.is_integer() and 1 <= c <= n:  # Check if c is an integer and within range
                count += 1  # Count valid (a, b, c)
    return count

# Example usage
print(countSquareSumTriples(5))  # Output: 2
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