LeetCode #473 — MEDIUM

Matchsticks to Square

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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:

Input: matchsticks = [3,3,3,3,4]
Output: false
Explanation: You cannot find a way to form a square with all the matchsticks.

Constraints:

1 <= matchsticks.length <= 15
1 <= matchsticks[i] <= 10⁸

Step 01

Brute Force Baseline

Problem summary: You are given an integer array matchsticks where matchsticks[i] is the length of the ith 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[1,1,2,2,2]

Example 2

[3,3,3,3,4]

Core Insight

What unlocks the optimal approach

Treat the matchsticks as an array. Can we split the array into 4 equal parts?
Every matchstick can belong to either of the 4 sides. We don't know which one. Maybe try out all options!
For every matchstick, we have to try out each of the 4 options i.e. which side it can belong to. We can make use of recursion for this.
We don't really need to keep track of which matchsticks belong to a particular side during recursion. We just need to keep track of the <b>length</b> of each of the 4 sides.
When all matchsticks have been used we simply need to see the length of all 4 sides. If they're equal, we have a square on our hands!

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #473: Matchsticks to Square
class Solution {
    public boolean makesquare(int[] matchsticks) {
        int s = 0, mx = 0;
        for (int v : matchsticks) {
            s += v;
            mx = Math.max(mx, v);
        }
        int x = s / 4, mod = s % 4;
        if (mod != 0 || x < mx) {
            return false;
        }
        Arrays.sort(matchsticks);
        int[] edges = new int[4];
        return dfs(matchsticks.length - 1, x, matchsticks, edges);
    }

    private boolean dfs(int u, int x, int[] matchsticks, int[] edges) {
        if (u < 0) {
            return true;
        }
        for (int i = 0; i < 4; ++i) {
            if (i > 0 && edges[i - 1] == edges[i]) {
                continue;
            }
            edges[i] += matchsticks[u];
            if (edges[i] <= x && dfs(u - 1, x, matchsticks, edges)) {
                return true;
            }
            edges[i] -= matchsticks[u];
        }
        return false;
    }
}

// Accepted solution for LeetCode #473: Matchsticks to Square
func makesquare(matchsticks []int) bool {
	s := 0
	for _, v := range matchsticks {
		s += v
	}
	if s%4 != 0 {
		return false
	}
	sort.Sort(sort.Reverse(sort.IntSlice(matchsticks)))
	edges := make([]int, 4)
	var dfs func(u, x int) bool
	dfs = func(u, x int) bool {
		if u == len(matchsticks) {
			return true
		}
		for i := 0; i < 4; i++ {
			if i > 0 && edges[i-1] == edges[i] {
				continue
			}
			edges[i] += matchsticks[u]
			if edges[i] <= x && dfs(u+1, x) {
				return true
			}
			edges[i] -= matchsticks[u]
		}
		return false
	}
	return dfs(0, s/4)
}

# Accepted solution for LeetCode #473: Matchsticks to Square
class Solution:
    def makesquare(self, matchsticks: List[int]) -> bool:
        def dfs(u):
            if u == len(matchsticks):
                return True
            for i in range(4):
                if i > 0 and edges[i - 1] == edges[i]:
                    continue
                edges[i] += matchsticks[u]
                if edges[i] <= x and dfs(u + 1):
                    return True
                edges[i] -= matchsticks[u]
            return False

        x, mod = divmod(sum(matchsticks), 4)
        if mod or x < max(matchsticks):
            return False
        edges = [0] * 4
        matchsticks.sort(reverse=True)
        return dfs(0)

// Accepted solution for LeetCode #473: Matchsticks to Square
impl Solution {
    pub fn makesquare(matchsticks: Vec<i32>) -> bool {
        let mut matchsticks = matchsticks;

        fn dfs(matchsticks: &Vec<i32>, edges: &mut [i32; 4], u: usize, x: i32) -> bool {
            if u == matchsticks.len() {
                return true;
            }
            for i in 0..4 {
                if i > 0 && edges[i - 1] == edges[i] {
                    continue;
                }
                edges[i] += matchsticks[u];
                if edges[i] <= x && dfs(matchsticks, edges, u + 1, x) {
                    return true;
                }
                edges[i] -= matchsticks[u];
            }
            false
        }

        let sum: i32 = matchsticks.iter().sum();
        if sum % 4 != 0 {
            return false;
        }
        matchsticks.sort_by(|x, y| y.cmp(x));
        let mut edges = [0; 4];

        dfs(&matchsticks, &mut edges, 0, sum / 4)
    }
}

// Accepted solution for LeetCode #473: Matchsticks to Square
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #473: Matchsticks to Square
// class Solution {
//     public boolean makesquare(int[] matchsticks) {
//         int s = 0, mx = 0;
//         for (int v : matchsticks) {
//             s += v;
//             mx = Math.max(mx, v);
//         }
//         int x = s / 4, mod = s % 4;
//         if (mod != 0 || x < mx) {
//             return false;
//         }
//         Arrays.sort(matchsticks);
//         int[] edges = new int[4];
//         return dfs(matchsticks.length - 1, x, matchsticks, edges);
//     }
// 
//     private boolean dfs(int u, int x, int[] matchsticks, int[] edges) {
//         if (u < 0) {
//             return true;
//         }
//         for (int i = 0; i < 4; ++i) {
//             if (i > 0 && edges[i - 1] == edges[i]) {
//                 continue;
//             }
//             edges[i] += matchsticks[u];
//             if (edges[i] <= x && dfs(u - 1, x, matchsticks, edges)) {
//                 return true;
//             }
//             edges[i] -= matchsticks[u];
//         }
//         return false;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.