<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 rec(int i,vector<int> &nums,int k,vector<int> &v,int s)
    {
        int n=nums.size();
        if(i==n)
        {
            for(int j=0;j<k;j++)
                if(v[j]!=v[0]) return false;
            return true;
        }
        bool ans=false;
        unordered_map<int,int> M;
        for(int j=0;j<k;j++)
        {
            if(M[v[j]]) continue;
            M[v[j]]=1;
            if(v[j]+nums[i]>s/k) continue;
            v[j]+=nums[i];
            ans|=rec(i+1,nums,k,v,s);
            v[j]-=nums[i];
            if(ans) return ans;
        }
        return ans;
    }

    bool makesquare(vector<int>& nums) {
        int k=4;
        vector<int> v(k,0);
        int s=0;
        if(nums.size()<k) return false;
        for(auto i:nums) s+=i;
        if(s%k!=0) return false;
        sort(nums.begin(),nums.end());
        reverse(nums.begin(),nums.end());
        for(auto i:nums)
            if(i>s/k) return false;
        return rec(0,nums,k,v,s);
    }
};
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