What is the time complexity of Perfect Squares?

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

What data structures are used to solve Perfect Squares?

This problem uses Math, Dynamic Programming, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Perfect Squares?

Insight 1: Legendre's three-square theorem can be used to determine if a number can be represented as the sum of three squares. This is useful to avoid unnecessary computations. Insight 2: Lagrange's four-square theorem states that every positive integer can be expressed as the sum of at most four perfect squares.

Which companies ask LeetCode 279. Perfect Squares in interviews?

This problem is frequently asked at Accenture, Citadel, Revolut, Walmart Labs, Yandex.

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Perfect Squares - LeetCode 279 Solution

Q: How to solve Perfect Squares?

The solution uses a combination of mathematical properties and a simple iterative approach to determine the minimum number of perfect squares. It first checks if 'n' is itself a perfect square. If not, it applies a transformation based on Legendre's three-square theorem and Lagrange's four-square theorem to potentially reduce the problem to finding the sum of at most three squares. Finally, it iterates through possible square numbers less than 'n' to determine if 'n' can be expressed as the sum of two squares. If none of these cases hold, it returns 3 based on Lagrange's theorem that every positive integer can be expressed as the sum of at most four squares.

Perfect Squares - Complete Solution Guide

Perfect Squares is LeetCode problem 279, 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 an integer n , return the least number of perfect square numbers that sum to n . A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1 , 4 , 9 , and 16 are perfect squares while 3 and 11 are not. Example 1: Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4. Example 2: Input: n = 13 Output: 2 Explanation: 13 = 4 + 9. Constraints: 1 <= n <= 10 4

Detailed Explanation

The problem asks us to find the minimum number of perfect squares (e.g., 1, 4, 9, 16, ...) that sum up to a given positive integer 'n'. For example, if n = 12, the answer is 3 because 12 = 4 + 4 + 4. If n = 13, the answer is 2 because 13 = 4 + 9.

Solution Approach

The solution uses a combination of mathematical properties and a simple iterative approach to determine the minimum number of perfect squares. It first checks if 'n' is itself a perfect square. If not, it applies a transformation based on Legendre's three-square theorem and Lagrange's four-square theorem to potentially reduce the problem to finding the sum of at most three squares. Finally, it iterates through possible square numbers less than 'n' to determine if 'n' can be expressed as the sum of two squares. If none of these cases hold, it returns 3 based on Lagrange's theorem that every positive integer can be expressed as the sum of at most four squares.

Step-by-Step Algorithm

Step 1: Check if 'n' is a perfect square. If it is, return 1.
Step 2: Reduce 'n' by repeatedly dividing it by 4 as long as it's divisible by 4. This is based on Legendre's three-square theorem.
Step 3: Check if the reduced 'n' is congruent to 7 modulo 8. If it is, return 4. This step utilizes Legendre's three-square theorem.
Step 4: Iterate through all possible perfect squares less than or equal to 'n'. For each square 'i*i', check if 'n - i*i' is also a perfect square. If it is, return 2.
Step 5: If none of the above conditions are met, return 3.

Key Insights

Insight 1: Legendre's three-square theorem can be used to determine if a number can be represented as the sum of three squares. This is useful to avoid unnecessary computations.
Insight 2: Lagrange's four-square theorem states that every positive integer can be expressed as the sum of at most four perfect squares.
Insight 3: Checking if a number is already a perfect square is the first optimization and is crucial for reducing the problem to simpler cases.

Complexity Analysis

Time Complexity: O(sqrt(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Breadth-First Search.

Companies

Asked at: Accenture, Citadel, Revolut, Walmart Labs, Yandex.

Perfect Squares | LeetCode 279

Medium

Math Dynamic Programming Breadth-First Search

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

Solution Code

Available in:

class Solution {
    public int numSquares(int n) {
        if ((int)Math.sqrt(n) * (int)Math.sqrt(n) == n) {
            return 1;
        }
        
        int temp_n = n;
        while (temp_n % 4 == 0) {
            temp_n /= 4;
        }
        if (temp_n % 8 == 7) {
            return 4;
        }
        
        int i = 1;
        while (i * i <= n) {
            int j = n - i * i;
            if ((int)Math.sqrt(j) * (int)Math.sqrt(j) == j) {
                return 2;
            }
            i += 1;
        }
        
        return 3;
    }
}

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Problem Info

Description

Problem 279: Perfect Squares - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Breadth-First Search

Companies

AccentureCitadelRevolutWalmart LabsYandex

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:279

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

5 steps

1
Step 1: Check if 'n' is a perfect square. If it is, return 1.
2
Step 2: Reduce 'n' by repeatedly dividing it by 4 as long as it's divisible by 4. This is based on Legendre's three-square theorem.
3
Step 3: Check if the reduced 'n' is congruent to 7 modulo 8. If it is, return 4. This step utilizes Legendre's three-square theorem.
4
Step 4: Iterate through all possible perfect squares less than or equal to 'n'. For each square 'i*i', check if 'n - i*i' is also a perfect square. If it is, return 2.
5
Step 5: If none of the above conditions are met, return 3.

💡

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(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the while loops where we repeatedly divide by 4, and where we iterate through possible square roots. Space complexity is constant because the algorithm only uses a few integer variables.

📖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

Dynamic Programming

A dynamic programming approach builds a table dp[i] representing the minimum number of perfect squares summing to i. dp[i] = min(dp[i-j*j] + 1) for all j such that j*j <= i.

Time: O(n * sqrt(n))Space: O(n)

Trade-offs: The dynamic programming approach is guaranteed to find the correct answer but may be slower than the provided solution for some inputs due to the larger time complexity. It has a larger space complexity. The provided solution leverages number theory properties, making it efficient. DP is preferred when number theory tricks don't immediately come to mind.

Breadth-First Search (BFS)

BFS can be used to explore the possible combinations of perfect squares. Start with 'n' and subtract perfect squares at each level. The first time you reach 0, the level represents the minimum number of perfect squares.

Time: O(n * sqrt(n)) in the worst case, though it can be faster in some cases.Space: O(n) due to the queue.

Trade-offs: BFS can also be used, but its time and space complexity is similar to or worse than dynamic programming. However, BFS may find the answer faster in some cases, as it stops when a solution is found rather than computing all possible solutions like dynamic programming.

✨

Key Insights

1Insight 1: Legendre's three-square theorem can be used to determine if a number can be represented as the sum of three squares. This is useful to avoid unnecessary computations.
2Insight 2: Lagrange's four-square theorem states that every positive integer can be expressed as the sum of at most four perfect squares.
3Insight 3: Checking if a number is already a perfect square is the first optimization and is crucial for reducing the problem to simpler cases.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly returns 1 because 1 is a perfect square.
Edge case 2: n = a perfect square (e.g., n = 4, 9, 16). The solution efficiently identifies this and returns 1.
Edge case 3: n is such that it can be expressed as the sum of two squares (e.g., n = 5, 10). The iterative checking in the solution identifies this and returns 2.
Edge case 4: n = a number that requires 4 perfect squares, like numbers congruent to 7 mod 8 after repeatedly dividing by 4. The solution correctly returns 4

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if 'n' is a perfect square at the beginning. This is a basic optimization that significantly reduces the problem's complexity when applicable.

✗Common mistake 2: Incorrectly applying Legendre's three-square theorem. Make sure the conditions (dividing by 4, checking modulo 8) are applied in the correct order.

✗Common mistake 3: Using an inefficient loop or method to determine if a number is a perfect square. Using `int(sqrt(n)) ** 2 == n` is an effective method in most languages for checking if n is a perfect square.

✅

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: Optimizing resource allocation in various computational problems. For example, minimizing the number of computational tasks (where each task represents a perfect square) required to achieve a certain goal (represented by 'n').

🏦Real-world scenario 2: This problem highlights the importance of number theory in optimization. In areas like cryptography or data compression, understanding number properties can lead to more efficient algorithms.

💡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 8, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Perfect Squares"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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