<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 dfs(vector<int> matchsticks, int target, vector<int> &sides, int idx){
        
        // base case
        if(idx == matchsticks.size()){
            if(sides[0] == sides[1] && sides[1] == sides[2] && sides[2] == sides[3]){
                return true;
            }
            return false;
        }
        
        for(int i = 0; i < 4; i++){
            if (sides[i] + matchsticks[idx] > target){  //optimization - 1
                continue;
            }
            int j = i - 1;  //optimization - 3
            while (j>=0){
                if(sides[j] == sides[i]){
                    break;
                }
                
                j -= 1;
            }
            
            if(j != -1){
                continue;
            }
            
            
            sides[i] += matchsticks[idx];
            
            if( dfs(matchsticks, target, sides, idx+1) ){
                return true;
            }
            sides[i] -= matchsticks[idx];
        }
        
        return false;
    
        
    }
    
    
    bool makesquare(vector<int>& matchsticks) {
        
        if(matchsticks.size() == 0){
            return false;
        }
        
        int sum = 0;
        for(int i = 0; i < matchsticks.size() ; i++){
            sum += matchsticks[i];
        }
        
        int target = sum / 4;
        vector<int> sides(4,0);
        sort(matchsticks.begin(), matchsticks.end(), greater<int>()); // optimization 2
        return dfs(matchsticks, target, sides, 0);
        
        
    }
};    
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