Count Square Sum Triples

Count Square Sum Triples - Problem

Math Enumeration Easy

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.

For example, if n = 5, the square triples are (3,4,5) and (4,3,5), so the answer is 2.

Input & Output

Example 1 — Basic Case

$ Input: n = 5

› Output: 2

💡 Note: The square triples are (3,4,5) and (4,3,5). Both satisfy 3² + 4² = 9 + 16 = 25 = 5² and all values are ≤ 5.

Example 2 — Small Range

$ Input: n = 3

› Output: 0

💡 Note: No valid square triples exist with all values ≤ 3. The smallest Pythagorean triple is (3,4,5) but 4 and 5 exceed the limit.

Example 3 — Larger Range

$ Input: n = 10

› Output: 4

💡 Note: Valid triples: (3,4,5), (4,3,5), (6,8,10), (8,6,10). Each satisfies the Pythagorean theorem with all values ≤ 10.

Constraints

1 ≤ n ≤ 250

Visualization

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Asked in

G Google 25 M Microsoft 20 A Apple 15

The key insight is using the Pythagorean theorem a² + b² = c² to identify valid square triples. Best approach is the two-loop optimization that calculates c from each (a,b) pair and validates it. Time: O(n²), Space: O(1)

Common Approaches

✓ Triple Loop Brute Force

⏱️ Time: O(n³) Space: O(1)

Try every possible combination of (a,b,c) where each is between 1 and n, then check if a² + b² = c².

Two Loop with Square Root

⏱️ Time: O(n²) Space: O(1)

For each pair (a,b), calculate c = √(a² + b²) and check if c is an integer within the range [1,n].

Triple Loop Brute Force — Algorithm Steps

Step 1: Loop through all values of a from 1 to n
Step 2: Loop through all values of b from 1 to n
Step 3: Loop through all values of c from 1 to n
Step 4: Check if a² + b² equals c²

Visualization

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Step-by-Step Walkthrough

Generate All Triples

Try every (a,b,c) where 1≤a,b,c≤n

Check Pythagorean

Test if a²+b²=c²

Count Valid

Increment counter for each valid triple

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    int count = 0;
    for (int a = 1; a <= n; a++) {
        for (int b = a; b <= n; b++) {
            for (int c = 1; c <= n; c++) {
                if (a * a + b * b == c * c) {
                    count++;
                }
            }
        }
    }
    return count;
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops each running n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to store counts

✓ Linear Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Triple Loop Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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