<p>You are given an integer array <code>matchsticks</code> where <code>matchsticks[i]</code> is the length of the <code>i<sup>th</sup></code> matchstick. You want to use <strong>all the matchsticks</strong> to make one square. You <strong>should not break</strong> any stick, but you can link them up, and each matchstick must be used <strong>exactly one time</strong>.</p>

<p>Return <code>true</code> if you can make this square and <code>false</code> otherwise.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/09/matchsticks1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> matchsticks = [1,1,2,2,2]
<strong>Output:</strong> true
<strong>Explanation:</strong> You can form a square with length 2, one side of the square came two sticks with length 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= matchsticks.length &lt;= 15</code></li>
	<li><code>1 &lt;= matchsticks[i] &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution {
public:
    bool makesquare(vector<int>& A) {
        long sideSum = accumulate(A.begin(), A.end(), 0L);
        int n = A.size();
        
        // Check if the total length can be evenly divided into 4 sides
        if (sideSum % 4 != 0 || *max_element(A.begin(), A.end()) > sideSum / 4) 
            return false;
        
        sideSum /= 4;
        sort(A.begin(), A.end(), greater<>()); // Sort in descending order
        
        // Array to store the current sum of each side
        long sides[4] = {};
        
        function<bool(int)> dfs = [&](int index) {
            if (index == n) return true; // All matchsticks used
            
            unordered_set<long> visited; // Track visited states to avoid redundant checks
            
            for (int i = 0; i < 4; ++i) {
                // Skip if adding the current matchstick exceeds the side length
                // or if the current side state has been visited
                if (A[index] + sides[i] > sideSum || visited.count(sides[i])) 
                    continue;
                
                visited.insert(sides[i]);
                sides[i] += A[index]; // Add matchstick to the current side
                
                if (dfs(index + 1)) return true; // Recurse to place the next matchstick
                
                sides[i] -= A[index]; // Backtrack if placement fails
            }
            
            return false; // No valid configuration found
        };
        
        return dfs(0); // Start recursion with the first matchstick
    }
};

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