#1925

Easy

LeetCode Problem Workspace

Count Square Sum Triples

Count Square Sum Triples asks to find the number of integer triples where a² + b² = c² for values between 1 and n.

Problem Statement

Given an integer n, your task is to find the number of square triples (a, b, c) such that a² + b² = c², where a, b, and c are integers and 1 <= a, b, c <= n.

For example, when n = 5, the square triples are (3,4,5) and (4,3,5), so the answer is 2. You need to count such square triples for any given n up to 250.

Examples

Example 1

Input: n = 5

Output: 2

The square triples are (3,4,5) and (4,3,5).

Example 2

Input: n = 10

Output: 4

The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).

Constraints

1 <= n <= 250

Solution Approach

Brute Force Enumeration

Start by iterating over all possible pairs (a, b) within the range [1, n]. For each pair, compute the sum a² + b², then check if the square root of this sum is an integer and whether it is less than or equal to n. If these conditions hold, count the pair as a valid square triple.

Optimized Square Check

Instead of checking for every possible pair (a, b), you can improve efficiency by limiting the number of checks using the properties of squares. By precomputing squares of numbers, the check becomes faster. You also avoid redundant calculations of square sums.

Efficient Pair Handling

Since (a, b) and (b, a) represent the same triple, ensure to count each valid pair only once. This avoids double-counting and ensures that the solution is both correct and efficient.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this approach depends on the number of pairs (a, b) to check. In the brute force case, this is O(n²), where n is the maximum value of a and b. The space complexity is O(1), as the solution uses constant space for computations. Optimizations can reduce the number of checks, but the time complexity remains dependent on the final approach.

What Interviewers Usually Probe

The candidate understands the concept of square triples and can break the problem into smaller, manageable parts.
The candidate demonstrates the ability to optimize brute force methods by applying mathematical properties.
The candidate shows attention to detail in ensuring that each valid pair is counted only once.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the fact that (a, b) and (b, a) represent the same triple, leading to double-counting.
Neglecting to check that the square root of a² + b² is an integer and not just a floating-point value.
Not handling the edge cases where n is small or large, leading to either incorrect answers or inefficient solutions.

Follow-up variants

Consider the problem for n up to a much larger value, e.g., 1000, and see how the solution needs to scale.
Modify the problem to only count distinct triples (a, b, c) without considering the order of a and b.
Extend the problem to check if a² + b² + c² = d² for four variables, increasing the complexity.

FAQ

What is a square triple?

A square triple (a, b, c) is a set of three integers where a² + b² = c², meaning the sum of the squares of a and b equals the square of c.

How do I optimize this problem?

You can optimize the brute force solution by precomputing squares, reducing the number of calculations, and eliminating redundant checks for pairs like (a, b) and (b, a).

How does GhostInterview assist with solving the problem?

GhostInterview helps by providing structured advice on mathematical enumeration, ensuring efficient handling of the problem's complexity and common pitfalls.

Why is it important to check if a² + b² is a perfect square?

Checking if a² + b² is a perfect square is crucial because the problem asks for valid triples where the sum of squares of a and b equals c², making the square root check essential.

What other problems are similar to Count Square Sum Triples?

Problems involving pairs, squares, and summation such as finding Pythagorean triples or those requiring enumeration over number ranges share similar techniques and patterns.

terminal

Solution

Solution 1: Enumeration

We enumerate $a$ and $b$ in the range $[1, n)$, then calculate $c = \sqrt{a^2 + b^2}$. If $c$ is an integer and $c \leq n$, then we have found a Pythagorean triplet, and we increment the answer by one.

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class Solution:
    def countTriples(self, n: int) -> int:
        ans = 0
        for a in range(1, n):
            for b in range(1, n):
                x = a * a + b * b
                c = int(sqrt(x))
                if c <= n and c * c == x:
                    ans += 1
        return ans

Continue Practicing

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