What is the time complexity of Sum of Square Numbers?

The optimal solution has a time complexity of O(sqrt(c)).

What data structures are used to solve Sum of Square Numbers?

This problem uses Math, Two Pointers, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Sum of Square Numbers?

Insight 1: The problem can be solved efficiently using a two-pointer approach, similar to the two-sum problem when dealing with a sorted array. Insight 2: The range of possible values for 'a' and 'b' is limited by the square root of 'c'. We can use this to constrain the search space and optimize the solution.

Which companies ask LeetCode 633. Sum of Square Numbers in interviews?

This problem is frequently asked at LinkedIn, Two Sigma.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Sum of Square Numbers - LeetCode 633 Solution

Sum of Square Numbers - Complete Solution Guide

Sum of Square Numbers is LeetCode problem 633, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a non-negative integer c , decide whether there're two integers a and b such that a 2 + b 2 = c . Example 1: Input: c = 5 Output: true Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: c = 3 Output: false Constraints: 0 <= c <= 2 31 - 1

Detailed Explanation

The problem asks us to determine if a given non-negative integer 'c' can be expressed as the sum of squares of two integers, 'a' and 'b'. In other words, we need to check if there exist integers 'a' and 'b' such that a2 + b2 = c. The input is a single integer 'c', and the output should be a boolean value: 'true' if such 'a' and 'b' exist, and 'false' otherwise. The constraint is that 0 <= c <= 231 - 1.

Solution Approach

The solution utilizes a two-pointer approach. One pointer, 'left', starts at 0, and the other pointer, 'right', starts at the square root of 'c'. We then iterate while 'left' is less than or equal to 'right'. In each iteration, we calculate the sum of squares of 'left' and 'right'. If the sum is equal to 'c', we return 'true'. If the sum is less than 'c', we increment 'left' to increase the sum. If the sum is greater than 'c', we decrement 'right' to decrease the sum. If the loop completes without finding a solution, we return 'false'.

Step-by-Step Algorithm

Step 1: Initialize two pointers, 'left' to 0 and 'right' to the integer square root of 'c'.
Step 2: Start a while loop that continues as long as 'left' is less than or equal to 'right'.
Step 3: Inside the loop, calculate the sum of squares: 'current_sum' = left * left + right * right.
Step 4: Compare 'current_sum' with 'c':
Step 5: If 'current_sum' == 'c', return 'true' because we found a solution.
Step 6: If 'current_sum' < 'c', increment 'left' to try a larger value for 'a'.
Step 7: If 'current_sum' > 'c', decrement 'right' to try a smaller value for 'b'.
Step 8: If the loop finishes without finding a solution (i.e., 'left' becomes greater than 'right'), return 'false'.

Key Insights

Insight 1: The problem can be solved efficiently using a two-pointer approach, similar to the two-sum problem when dealing with a sorted array.
Insight 2: The range of possible values for 'a' and 'b' is limited by the square root of 'c'. We can use this to constrain the search space and optimize the solution.
Insight 3: The problem can be also solved using Binary Search algorithm.

Complexity Analysis

Time Complexity: O(sqrt(c))

Space Complexity: O(1)

Topics

This problem involves: Math, Two Pointers, Binary Search.

Companies

Asked at: LinkedIn, Two Sigma.

Sum of Square Numbers | LeetCode 633

Medium

Math Two Pointers Binary Search

Time: O(sqrt(c))Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean judgeSquareSum(int c) {
        int left = 0;
        int right = (int)Math.sqrt(c);

        while (left <= right) {
            int current_sum = left * left + right * right;
            if (current_sum == c) {
                return true;
            } else if (current_sum < c) {
                left += 1;
            } else {
                right -= 1;
            }
        }
        
        return false;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 633: Sum of Square Numbers - Detailed explanation and solution approaches.

Topics

Math Two Pointers Binary Search

Companies

LinkedInTwo Sigma

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:633

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 two pointers, 'left' to 0 and 'right' to the integer square root of 'c'.
2
Step 2: Start a while loop that continues as long as 'left' is less than or equal to 'right'.
3
Step 3: Inside the loop, calculate the sum of squares: 'current_sum' = left * left + right * right.
4
Step 4: Compare 'current_sum' with 'c':
5
Step 5: If 'current_sum' == 'c', return 'true' because we found a solution.
6
Step 6: If 'current_sum' < 'c', increment 'left' to try a larger value for 'a'.
7
Step 7: If 'current_sum' > 'c', decrement 'right' to try a smaller value for 'b'.
8
Step 8: If the loop finishes without finding a solution (i.e., 'left' becomes greater than 'right'), return 'false'.

💡

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(sqrt(c))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of iterations of the while loop. Because the range of 'left' and 'right' is determined by the square root of 'c', the number of steps will be at most proportional to sqrt(c). The space complexity is constant because we only use a fixed number of variables, independent of the input size.

📖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

Binary Search

For each value of 'a' from 0 to sqrt(c), check if c - a2 is a perfect square using binary search.

Time: O(sqrt(c) * log(c))Space: O(1)

Trade-offs: While Binary Search can be used, the two-pointer approach is generally preferred due to its simpler implementation and better time complexity (O(sqrt(c)) vs. O(sqrt(c) * log(c))

✨

Key Insights

1Insight 1: The problem can be solved efficiently using a two-pointer approach, similar to the two-sum problem when dealing with a sorted array.
2Insight 2: The range of possible values for 'a' and 'b' is limited by the square root of 'c'. We can use this to constrain the search space and optimize the solution.
3Insight 3: The problem can be also solved using Binary Search algorithm.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: c = 0. The solution correctly returns 'true' because 02 + 02 = 0.
Edge case 2: c = 1. The solution correctly returns 'true' because 12 + 02 = 1.
Edge case 3: Large values of c close to 231 - 1. The algorithm handles this correctly as it calculates intermediate results as 64-bit integers to prevent overflow (CPP, C), or the languages themselves handle large values correctly (Python, Java).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. When calculating 'left * left + right * right', the result can exceed the maximum value of a 32-bit integer if 'c' is large. The C and CPP solutions use 'long long' to prevent this. The Java solution explicitly uses `int` and the test cases are designed so that left * left + right * right is still less than Integer.MAX_VALUE. Python handles large numbers automatically.

✗Common mistake 2: Incorrect loop condition. Using 'left < right' instead of 'left <= right' in the while loop can lead to missing some valid solutions, especially when a = b.

✅

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: Pythagorean theorem applications in geometry and physics, where determining if a triangle with a given hypotenuse can be formed with integer sides is equivalent to solving this problem.

🏦Real-world scenario 2: Cryptography, specifically in number theory related problems where finding sums of squares is relevant to certain encryption schemes.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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: September 2, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Sum of Square Numbers"?

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

🤖 Need help with this problem?