How do you solve Count Square Sum Triples on LeetCode?

Count Square Sum Triples (LeetCode #1925) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n√n) time complexity and O(1) space complexity. Key insight: Square triples are symmetric; (a, b, c) and (b, a, c) count as separate valid triples.

What is the brute force solution for Count Square Sum Triples?

The brute force approach for Count Square Sum Triples is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of integers (a, b) and calculating c using the equation a² + b² = c². If c is an integer and within the bounds, we count it as a valid square triple. The optimal Optimal Solution approach improves this to O(n√n).

What algorithm or data structure is used to solve Count Square Sum Triples?

Count Square Sum Triples uses the following concepts: Math, Enumeration. The recommended patterns to study are: Mathematical Enumeration, Two Pointers.

What are the interview tips for Count Square Sum Triples?

Always clarify the constraints and edge cases before jumping into coding. Discuss your thought process and approach with the interviewer to get feedback. Write clean and concise code; avoid unnecessary complexity.

What are common mistakes when solving Count Square Sum Triples?

Forgetting to check if b is within the bounds (1 to n). Not considering the symmetry in (a, b) pairs.

#1925

Count Square Sum Triples

Easy

MathEnumerationMathematical EnumerationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n√n)
Space	O(1)	O(1)

💡

Intuition

Time O(n√n)Space O(1)

Instead of checking every pair (a, b), we can iterate through possible values of c and find pairs (a, b) that satisfy the equation a² + b² = c². This reduces the number of iterations significantly.

⚙️

Algorithm

6 steps

1Step 1: Initialize a counter to zero.
2Step 2: Iterate over all values of c from 1 to n.
3Step 3: For each c, iterate over possible values of a from 1 to c (since a must be less than or equal to c).
4Step 4: Calculate b as sqrt(c² - a²).
5Step 5: Check if b is an integer and if b <= n. If true, increment the counter.
6Step 6: Return the counter.

solution.py14 lines

1# Full working Python code
2import math
3
4def countSquareSumTriples(n):
5    count = 0
6    for c in range(1, n + 1):
7        for a in range(1, c + 1):
8            b = math.sqrt(c * c - a * a)
9            if b.is_integer() and b <= n:
10                count += 1
11    return count
12
13# Example usage
14print(countSquareSumTriples(5))  # Output: 2

ℹ

Complexity note: The time complexity is O(n√n) because for each value of c, we iterate up to c, and calculating b involves a square root operation. The space complexity is O(1) since we are using a constant amount of space.

1Square triples are symmetric; (a, b, c) and (b, a, c) count as separate valid triples.
2The maximum value of c determines the range of possible a and b values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.