<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>& matchsticks) {
        int sum = accumulate(matchsticks.begin(), matchsticks.end(), 0);
        if(sum % 4) return false;
        int side = sum / 4;

        
        vector<unordered_set<unsigned long>> memo(4);

        function<bool(bitset<15>, int)> dfs =
        [&dfs, matchsticks, side, &memo](bitset<15> used, int completed) -> bool {
            // If we have built all four sides and used all matchsticks, it is possible
            // Otherwise, no
            if(completed == 4) return used.count() == matchsticks.size();

            if(memo[completed].count(used.to_ulong())) return false;

            int built_perimeter = 0;
            for(int i = 0; i < 15; i++) if(used[i]) built_perimeter += matchsticks[i];

            // Compute how much we have left for this side
            int this_side = built_perimeter - completed * side;
            int remainder = side - this_side;

            // Try each
            for(int i = 0; i < matchsticks.size(); i++) {
                if(used[i]) continue;
                int matchstick = matchsticks[i];
                if(matchstick > remainder) continue;

                used[i] = true;
                if(dfs(used, matchstick == remainder ? completed + 1 : completed)) return true;
                used[i] = false;
            }

            memo[completed].insert(used.to_ulong());
            return false;
        };

        bitset<15> used;
        return dfs(used, 0);
    }
};
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