<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 {
    map<pair<int,int>,bool> dp;
public:
    bool makesquare(vector<int>& matchsticks) {
        int n=matchsticks.size();
        if(n<4) return 0;

        int sum=0;
        for(int i=0;i<n;i++)
        {
            sum+=matchsticks[i];
        }

        if(sum%4) return 0;

        return f(matchsticks,(1<<15)-1,0,sum/4);
    }

    bool f(vector<int> &a, int masks, int side, int t)
    {
        int sum=0;
        int n=a.size();

        for(int i=n-1;i>=0;i--)
        {
            if(((1<<i)&masks)==0) sum+=a[n-1-i];
        }

        if(sum && sum%t==0) side++;

        if(side==3) return 1;

        if(dp.find({masks,side})!=dp.end()) return dp[{masks,side}];
        
        int rem=t-(sum)%t;

        for(int i=n-1;i>=0;i--)
        {
            if(a[n-1-i]<=rem && (masks&(1<<i)))
            {
                if(f(a,masks^(1<<i),side,t))
                {
                    return dp[{masks,side}]=1;
                }
            }
        }

        return dp[{masks,side}]=0;
    }
};
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