What is the time complexity of Count Square Sum Triples?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Count Square Sum Triples?

This problem uses Math, Enumeration. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Square Sum Triples?

Insight 1: Brute-force approach is feasible due to the constraint 1 <= n <= 250. A nested loop checking all possible combinations of a, b, and c is computationally manageable within this range. Insight 2: Efficiently checking for perfect squares. Instead of iterating through all possible values of 'c', calculating c = sqrt(a*a + b*b) and then checking if c*c == a*a + b*b is a more efficient approach. This avoids unnecessary iterations.

Which companies ask LeetCode 1925. Count Square Sum Triples in interviews?

This problem is frequently asked at Qualtrics.

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Count Square Sum Triples - LeetCode 1925 Solution

Q: How to solve Count Square Sum Triples?

The provided solutions employ a brute-force approach. They iterate through all possible pairs (a, b) within the range [1, n]. For each pair, they calculate a² + b² and check if the result is a perfect square. If it is, and the square root (c) is also within the range [1, n], the count of triples is incremented. This approach systematically checks all possible combinations to find the valid Pythagorean triples.

Q: Which companies ask LeetCode 1925. Count Square Sum Triples in interviews?

This problem is frequently asked at Qualtrics.

Count Square Sum Triples - Complete Solution Guide

Count Square Sum Triples is LeetCode problem 1925, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A square triple (a,b,c) is a triple where a , b , and c are integers and a 2 + b 2 = c 2 . 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

Detailed Explanation

The problem asks you to find the number of Pythagorean triples (a, b, c) where a, b, and c are integers between 1 and n (inclusive), and a² + b² = c². A Pythagorean triple is a set of three integers that satisfy the Pythagorean theorem (a² + b² = c²). The problem emphasizes that a, b, and c must all be within the range [1, n].

Solution Approach

The provided solutions employ a brute-force approach. They iterate through all possible pairs (a, b) within the range [1, n]. For each pair, they calculate a² + b² and check if the result is a perfect square. If it is, and the square root (c) is also within the range [1, n], the count of triples is incremented. This approach systematically checks all possible combinations to find the valid Pythagorean triples.

Step-by-Step Algorithm

Step 1: Initialize a counter 'count' to 0.
Step 2: Iterate through all possible values of 'a' from 1 to n using a nested loop.
Step 3: For each value of 'a', iterate through all possible values of 'b' from 1 to n.
Step 4: Calculate c² = a² + b².
Step 5: Calculate c = sqrt(c²). Check if c is an integer (c² is a perfect square) using c * c == c_squared.
Step 6: Check if c is within the range [1, n].
Step 7: If both conditions in steps 5 and 6 are true, increment 'count'.
Step 8: After iterating through all pairs (a, b), return 'count'.

Key Insights

Insight 1: Brute-force approach is feasible due to the constraint 1 <= n <= 250. A nested loop checking all possible combinations of a, b, and c is computationally manageable within this range.
Insight 2: Efficiently checking for perfect squares. Instead of iterating through all possible values of 'c', calculating c = sqrt(a*a + b*b) and then checking if c*c == a*a + b*b is a more efficient approach. This avoids unnecessary iterations.
Insight 3: Avoiding unnecessary calculations. The provided Python and Java solutions are optimized by not including a third nested loop for 'c', significantly improving the efficiency compared to the C++ solution which does.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Math, Enumeration.

Companies

Asked at: Qualtrics.

Count Square Sum Triples | LeetCode 1925

Easy

Math Enumeration

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countTriples(int n) {
        int count = 0;
        for (int a = 1; a <= n; a++) {
            for (int b = 1; b <= n; b++) {
                int cSquared = a * a + b * b;
                int c = (int) Math.sqrt(cSquared);
                if (c * c == cSquared && c <= n) {
                    count++;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 1925: Count Square Sum Triples - Detailed explanation and solution approaches.

Topics

Math Enumeration

Companies

Qualtrics

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1925

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a counter 'count' to 0.
2
Step 2: Iterate through all possible values of 'a' from 1 to n using a nested loop.
3
Step 3: For each value of 'a', iterate through all possible values of 'b' from 1 to n.
4
Step 4: Calculate c² = a² + b².
5
Step 5: Calculate c = sqrt(c²). Check if c is an integer (c² is a perfect square) using c * c == c_squared.
6
Step 6: Check if c is within the range [1, n].
7
Step 7: If both conditions in steps 5 and 6 are true, increment 'count'.
8
Step 8: After iterating through all pairs (a, b), return 'count'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n²) because of the nested loops iterating through all possible pairs of (a, b). The space complexity is O(1) because the algorithm uses a constant amount of extra space regardless of the input size 'n'. The C++ solution has a time complexity of O(n^3) due to the third nested loop.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Optimized Brute Force

Instead of checking all possible values of c, directly check if a^2 + b^2 is a perfect square.

Time: O(n^2)Space: O(1)

Trade-offs: This approach is slightly more efficient than the exhaustive search across c, as it reduces unnecessary calculations.

✨

Key Insights

1Insight 1: Brute-force approach is feasible due to the constraint 1 <= n <= 250. A nested loop checking all possible combinations of a, b, and c is computationally manageable within this range.
2Insight 2: Efficiently checking for perfect squares. Instead of iterating through all possible values of 'c', calculating c = sqrt(a*a + b*b) and then checking if c*c == a*a + b*b is a more efficient approach. This avoids unnecessary iterations.
3Insight 3: Avoiding unnecessary calculations. The provided Python and Java solutions are optimized by not including a third nested loop for 'c', significantly improving the efficiency compared to the C++ solution which does.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 1. The solution correctly handles this case, returning 0 as there are no triples possible.
Edge case 2: n = any value resulting in no triples. The algorithm correctly handles this by returning 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly checking for perfect squares. Failing to properly check if c*c == a*a + b*b, resulting in incorrect counts.

✗Common mistake 2: Forgetting to check the range of c (1 <= c <= n). This leads to incorrect inclusion of triples where c is outside the specified bounds.

✗Common mistake 3: Using inefficient nested loops in C++ solution which is O(n^3) instead of O(n^2)

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: In computer graphics, Pythagorean triples are used to generate points on a circle or sphere efficiently.

🏦Real-world scenario 2: Number theory applications, specifically in cryptography and certain types of code breaking and secure communication protocols, which need efficient checks for Pythagorean triples.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Math. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

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Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Square Sum Triples"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether enumeration techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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