<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>& v) {
        
        int tot = accumulate(begin(v) , end(v) , 0);
        
        if(tot%4 or v.size() < 4)return 0;
        
        vector<int>poss;
        
        int n = v.size();
        
        vector<int>ids(n);
        iota(begin(ids) , end(ids) , 0);
        
        function<int(int , vector<int> , int)>isPoss = [&](int target , vector<int>ids , int mode)-> int {
            
                int m = ids.size();
                for(int i = 0 ; i < (1<<m) ; ++i){
                    int sum = 0;
                    for(int j = 0 ; j < m ; ++j){
                        if(i&(1<<j))sum += v[mode ? ids[j] : n - 1 - ids[j]];
                    }
                    
                    if(sum == target){
                        if(mode) return 1;
                        else poss.push_back(i);
                    }
                }
                
                return 0;
        };
        
        isPoss(tot/2 , ids , 0);
        
        for(int&mask : poss){
            
            vector<int>s1 , s2;
            
            for(int i = 0 ; i < n ; ++i){
                if(mask&(1<<i)) s1.push_back(n-1-i);
                else s2.push_back(n-1-i);
            }
            
            int ok1 = isPoss(tot/4 , s1 , 1);
            int ok2 = isPoss(tot/4 , s2 , 1);
            
            if(ok1 and ok2) return 1;
        }
        
        return 0;
    }
};
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