<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 n = 0;
    int side = 0;
    set<vector<int> > dp[20];

    bool isSquarePossible(vector<int> m, int x, int a, int b, int c, int d) {
        if (x==n) {
            if (a==side && b==side && c==side && d==side)
            return true;
            return false;
        }

        if (a>side || b>side || c>side || d>side)
            return false;
        
        // choose each side one by one
        vector<int> v = {a+m[x],b,c,d};
        sort(v.begin(),v.end());
        if (dp[x+1].find(v) == dp[x+1].end() && isSquarePossible(m,x+1,a+m[x],b,c,d))
            return true;
        
        v = {a,b+m[x],c,d};
        sort(v.begin(),v.end());
        if (dp[x+1].find(v) == dp[x+1].end() && isSquarePossible(m, x+1,a,b+m[x],c,d))
            return true;

        v = {a,b,c+m[x],d};
        sort(v.begin(),v.end());
        if (dp[x+1].find(v) == dp[x+1].end() && isSquarePossible(m, x+1,a,b,c+m[x],d))
            return true;

        v = {a,b,c,d+m[x]};
        sort(v.begin(),v.end());
        if (dp[x+1].find(v) == dp[x+1].end() && isSquarePossible(m, x+1,a,b,c,d+m[x]))
            return true;
        
        v = {a,b,c,d};
        sort(v.begin(),v.end());
        dp[x].insert(v);
        return false;
    }

    bool makesquare(vector<int>& m) {
        n=m.size();
        for(auto u:m) {
            side += u;
        }
        if (side%4 !=0)
            return false;
        side/=4;
        sort(m.begin(),m.end());
        reverse(m.begin(),m.end());

        return isSquarePossible(m,0, 0,0,0,0);
    }
};
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