What is the time complexity of Matchsticks to Square?

The optimal solution has a time complexity of O(4^N).

What data structures are used to solve Matchsticks to Square?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Matchsticks to Square?

Insight 1: The sum of all matchstick lengths must be divisible by 4 for a square to be possible. This gives us the side length of the square. Insight 2: Backtracking is a suitable approach to explore all possible combinations of assigning matchsticks to the four sides of the square.

Which companies ask LeetCode 473. Matchsticks to Square in interviews?

This problem is frequently asked at PhonePe, eBay.

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

Matchsticks to Square - LeetCode 473 Solution

Matchsticks to Square - Complete Solution Guide

Matchsticks to Square is LeetCode problem 473, 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

You are given an integer array matchsticks where matchsticks[i] is the length of the i th matchstick. You want to use all the matchsticks to make one square. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time . Return true if you can make this square and false otherwise. Example 1: Input: matchsticks = [1,1,2,2,2] Output: true Explanation: You can form a square with length 2, one side of the square came two sticks with length 1. Example 2:

Detailed Explanation

The problem asks us to determine whether it's possible to form a square using a given set of matchsticks, without breaking any of them. Each matchstick must be used exactly once, and they can be joined at their ends. The input is an array of integers representing the lengths of the matchsticks. The output should be `true` if a square can be formed, and `false` otherwise. The constraints are that the number of matchsticks is between 1 and 15, and the length of each matchstick is between 1 and 10^8.

Solution Approach

The solution uses a backtracking algorithm to explore all possible combinations of assigning matchsticks to the four sides of the square. It first checks if the sum of the matchsticks is divisible by 4. If not, it returns `false`. Then, it calculates the target length of each side of the square. The matchsticks are sorted in descending order to prioritize larger sticks. The `backtrack` function recursively tries to add each matchstick to one of the four sides. If a matchstick can be added without exceeding the target length, the function recursively calls itself with the next matchstick. If the base case is reached where all matchsticks have been assigned, it means a square can be formed, and it returns `true`. Backtracking ensures exploration of all valid combinations by adding and removing matchsticks from the current side.

Step-by-Step Algorithm

Step 1: Calculate the total length of all matchsticks.
Step 2: Check if the total length is divisible by 4. If not, return `false`.
Step 3: Calculate the target length for each side of the square (total length / 4).
Step 4: Sort the matchsticks array in descending order.
Step 5: Define a recursive `backtrack` function that takes the current index as input.
Step 6: In the `backtrack` function, if the index reaches the end of the array (all matchsticks have been used), return `true`.
Step 7: Iterate through the four sides of the square.
Step 8: For each side, check if adding the current matchstick's length to the side's current length exceeds the target length, and skips duplicate calculations.
Step 9: If the matchstick can be added to the side, add it and recursively call `backtrack` with the next index.
Step 10: If the recursive call returns `true`, return `true` immediately.
Step 11: If the recursive call returns `false`, backtrack by subtracting the matchstick's length from the side and continue to the next side.
Step 12: If none of the sides can accommodate the current matchstick, return `false`.
Step 13: Call the `backtrack` function with the initial index of 0 and return its result.

Key Insights

Insight 1: The sum of all matchstick lengths must be divisible by 4 for a square to be possible. This gives us the side length of the square.
Insight 2: Backtracking is a suitable approach to explore all possible combinations of assigning matchsticks to the four sides of the square.
Insight 3: Sorting the matchsticks in descending order can improve efficiency by pruning the search space early. If a larger matchstick cannot fit on any of the sides, we can avoid exploring further combinations from that point.

Complexity Analysis

Time Complexity: O(4^N)

Space Complexity: O(N)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask.

Companies

Asked at: PhonePe, eBay.

Matchsticks to Square | LeetCode 473

Medium

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Time: O(4^N)Space: O(N)

Solution Code

Available in:

class Solution {
    public boolean makesquare(int[] matchsticks) {
        int total_len = 0;
        for (int matchstick : matchsticks) {
            total_len += matchstick;
        }
        if (total_len % 4 != 0) {
            return false;
        }

        int target = total_len / 4;
        int n = matchsticks.length;
        
        java.util.Arrays.sort(matchsticks);
        
        // Reverse the array to mimic the Python's sort(reverse=True)
        for (int i = 0; i < n / 2; i++) {
            int temp = matchsticks[i];
            matchsticks[i] = matchsticks[n - 1 - i];
            matchsticks[n - 1 - i] = temp;
        }
        
        if (matchsticks[0] > target) {
            return false;
        }
        
        int[] sides = new int[4];
        
        return backtrack(0, matchsticks, sides, target, n);
    }
    
    private boolean backtrack(int index, int[] matchsticks, int[] sides, int target, int n) {
        if (index == n) {
            return true;
        }
        
        for (int i = 0; i < 4; i++) {
            boolean is_duplicate_try = false;
            for (int j = 0; j < i; j++) {
                if (sides[j] == sides[i]) {
                    is_duplicate_try = true;
                    break;
                }
            }
            if (is_duplicate_try) {
                continue;
            }

            if (sides[i] + matchsticks[index] <= target) {
                sides[i] += matchsticks[index];
                if (backtrack(index + 1, matchsticks, sides, target, n)) {
                    return true;
                }
                sides[i] -= matchsticks[index];
            }
        }
        
        return false;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 473: Matchsticks to Square - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Companies

PhonePeeBay

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:473

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

13 steps

1
Step 1: Calculate the total length of all matchsticks.
2
Step 2: Check if the total length is divisible by 4. If not, return `false`.
3
Step 3: Calculate the target length for each side of the square (total length / 4).
4
Step 4: Sort the matchsticks array in descending order.
5
Step 5: Define a recursive `backtrack` function that takes the current index as input.
6
Step 6: In the `backtrack` function, if the index reaches the end of the array (all matchsticks have been used), return `true`.
7
Step 7: Iterate through the four sides of the square.
8
Step 8: For each side, check if adding the current matchstick's length to the side's current length exceeds the target length, and skips duplicate calculations.
9
Step 9: If the matchstick can be added to the side, add it and recursively call `backtrack` with the next index.
10
Step 10: If the recursive call returns `true`, return `true` immediately.
11
Step 11: If the recursive call returns `false`, backtrack by subtracting the matchstick's length from the side and continue to the next side.
12
Step 12: If none of the sides can accommodate the current matchstick, return `false`.
13
Step 13: Call the `backtrack` function with the initial index of 0 and return its result.

💡

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(4^N)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is exponential because, for each matchstick, we have 4 choices of which side to try to add it to. This results in a branching factor of 4. The space complexity is determined by the call stack of the recursive function. In the worst case, we might have N calls on the stack at once, which is O(N).

📖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 with Bitmasking

This approach uses a bitmask to represent which matchsticks have been used and stores the intermediate sums in a DP table. This can reduce the number of redundant calculations.

Time: O(N * 4^N), where N is the number of matchsticksSpace: O(4^N), where N is the number of matchsticks

Trade-offs: DP with bitmasking can be slightly more efficient than backtracking in some cases, especially when there are many overlapping subproblems. However, the space complexity is also exponential, and for smaller values of N, the overhead of DP might outweigh the benefits. Backtracking is generally easier to implement and understand.

✨

Key Insights

1Insight 1: The sum of all matchstick lengths must be divisible by 4 for a square to be possible. This gives us the side length of the square.
2Insight 2: Backtracking is a suitable approach to explore all possible combinations of assigning matchsticks to the four sides of the square.
3Insight 3: Sorting the matchsticks in descending order can improve efficiency by pruning the search space early. If a larger matchstick cannot fit on any of the sides, we can avoid exploring further combinations from that point.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: If the sum of matchsticks is not divisible by 4. The solution correctly returns `false` immediately.
Edge case 2: If the length of the longest matchstick is greater than the target side length. The solution correctly returns `false` after sorting the array.
Edge case 3: Empty input array. While the problem states that the length is at least 1, checking for an empty array `matchsticks` is an easy addition that improves robustness and doesn't affect complexity.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the sum of matchsticks is divisible by 4 before proceeding with the algorithm. This can lead to incorrect results and wasted computation.

✗Common mistake 2: Not sorting the matchsticks in descending order. Sorting helps prune the search space early and improves the efficiency of the backtracking algorithm. Without sorting, the algorithm might explore many unnecessary combinations.

✗Common mistake 3: Not correctly backtracking when a branch of the search tree fails. Failing to reset the sides array after an unsuccessful attempt to add a matchstick will lead to incorrect results.

✅

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: Resource allocation. This algorithm can be adapted to solve resource allocation problems where the goal is to divide resources (e.g., tasks, budget, manpower) into equal groups or categories.

🏦Real-world scenario 2: Load balancing. The core idea of partitioning items into equal groups is similar to load balancing in distributed systems, where the goal is to distribute workloads across multiple servers or nodes to ensure even utilization and prevent bottlenecks.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 22, 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 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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 "Matchsticks to Square"?

For this medium-level Array 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.

🤖 Need help with this problem?