<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:
    int lim;
    bool makesquare(vector<int>ms,vector<int>&dir,int n,int i){
        
        if(i == n){
            return true;
        }

        for(int j=0;j<4;j++){
            if(ms[i]+dir[j] <= lim){
                dir[j]+=ms[i];
                if(makesquare(ms,dir,n,i+1)) return true;
                dir[j]-=ms[i];
            }
            if(dir[j]==0) break;
        }
        return false;
       
    }
    bool makesquare(vector<int>& ms) {
        int sum=0;
        int tot = accumulate(ms.begin(),ms.end(),sum);
        if(tot%4 != 0) return false;
        lim = tot/4;
        vector<int>dir(4,0);

        sort(ms.rbegin(),ms.rend());;
        return makesquare(ms,dir,ms.size(),0);
    }
};
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