<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 n = matchsticks.size();
        int tot = 0;
        vector<int>sum(1<<n);
        vector<int>dp(1<<n,-69);
        dp[0] = 0;
        for(int nxt: matchsticks){
            tot+=nxt;
        }
        if(tot%4 != 0){
            return false;
        }
        tot/=4;
        for(int i = 1; i<(1<<n); i++){
            for(int j = 0; j<n; j++){
                if(i&(1<<j)){
                    sum[i]+=matchsticks[j];
                }
            }
        }
        for(int i = 1; i<(1<<n); i++){
            for(int j = i; j>0; j = (j-1)&i){
                if(sum[j] == tot)dp[i] = max(dp[i],dp[i^j]+1);
            }
        }
        return dp[(1<<n)-1] == 4;
    }
};
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