<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 {
    vector<vector<int>> dp = vector<vector<int>> (3, vector<int>(1024 * 32, -1));

    bool topdown(int xi, int used, vector<int>& matchsticks, int target) {
        if (xi == 3)
            return true;

        if (dp[xi][used] != -1)
            return dp[xi][used] == 1;

        bool ans = false;
        for (int to_use = 0 ; to_use < (1 << (int)matchsticks.size()) && !ans ; to_use++) {
            if ((to_use & used) != 0)
                continue;

            int sum_xi = 0;
            for (int i = 0 ; i < matchsticks.size() ; i++)
                if (((to_use >> i) & 1) == 1)
                    sum_xi += matchsticks[i];

            if (sum_xi != target)
                continue;

            ans |= topdown(xi+1, used ^ to_use, matchsticks, target);
        }

        dp[xi][used] = (ans)? 1:0;
        return ans;
    }

public:
    bool makesquare(vector<int>& matchsticks) {
        int sum = accumulate(matchsticks.begin(), matchsticks.end(), 0);
        if (sum % 4 != 0)
            return false;

        return topdown(0, 0, matchsticks, sum / 4);
    }
};
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