<p>You are given two groups of points where the first group has <code>size<sub>1</sub></code> points, the second group has <code>size<sub>2</sub></code> points, and <code>size<sub>1</sub> &gt;= size<sub>2</sub></code>.</p>

<p>The <code>cost</code> of the connection between any two points are given in an <code>size<sub>1</sub> x size<sub>2</sub></code> matrix where <code>cost[i][j]</code> is the cost of connecting point <code>i</code> of the first group and point <code>j</code> of the second group. The groups are connected if <strong>each point in both groups is connected to one or more points in the opposite group</strong>. In other words, each point in the first group must be connected to at least one point in the second group, and each point in the second group must be connected to at least one point in the first group.</p>

<p>Return <em>the minimum cost it takes to connect the two groups</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/03/ex1.jpg" style="width: 322px; height: 243px;" />
<pre>
<strong>Input:</strong> cost = [[15, 96], [36, 2]]
<strong>Output:</strong> 17
<strong>Explanation</strong>: The optimal way of connecting the groups is:
1--A
2--B
This results in a total cost of 17.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/03/ex2.jpg" style="width: 322px; height: 403px;" />
<pre>
<strong>Input:</strong> cost = [[1, 3, 5], [4, 1, 1], [1, 5, 3]]
<strong>Output:</strong> 4
<strong>Explanation</strong>: The optimal way of connecting the groups is:
1--A
2--B
2--C
3--A
This results in a total cost of 4.
Note that there are multiple points connected to point 2 in the first group and point A in the second group. This does not matter as there is no limit to the number of points that can be connected. We only care about the minimum total cost.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> cost = [[2, 5, 1], [3, 4, 7], [8, 1, 2], [6, 2, 4], [3, 8, 8]]
<strong>Output:</strong> 10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>size<sub>1</sub> == cost.length</code></li>
	<li><code>size<sub>2</sub> == cost[i].length</code></li>
	<li><code>1 &lt;= size<sub>1</sub>, size<sub>2</sub> &lt;= 12</code></li>
	<li><code>size<sub>1</sub> &gt;= size<sub>2</sub></code></li>
	<li><code>0 &lt;= cost[i][j] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int group1Size, group2Size;
    int dp[13][13][(1<<13)];  
    int solve(int ind1, int ind2, int matchedMask, vector<vector<int>>& cost) 
    {
        // Base case: All elements from group 1 have been processed
        if (ind1 == group1Size) 
        {
            return (matchedMask == (1 << group2Size) - 1) ? 0 : 1e9;
        }
        
        if (ind2 >= group2Size) 
        {
            return 1e9; 
        }

        if (dp[ind1][ind2][matchedMask] != -1) 
        {
            return dp[ind1][ind2][matchedMask];
        }

        // Option 1: Match element ind1 from group 1 with ind2 from group 2, then move to the next element in group 1
        int option1 = cost[ind1][ind2] + solve(ind1 + 1, 0, matchedMask | (1 << ind2), cost);
        
        // Option 2: Match element ind1 from group 1 with ind2 from group 2, but stay on the same element in group 1 and move to the next element in group 2
        int option2 = cost[ind1][ind2] + solve(ind1, ind2 + 1, matchedMask | (1 << ind2), cost);
        
        // Option 3: Skip matching the current element from group 1 with ind2 in group 2, and move to the next element in group 2
        int option3 = solve(ind1, ind2 + 1, matchedMask, cost);
        
        return dp[ind1][ind2][matchedMask] = min({option1, option2, option3});
    }

    int connectTwoGroups(vector<vector<int>>& cost) 
    {
        group1Size = cost.size();    
        group2Size = cost[0].size(); 
        int matchedMask = 0;         // Bitmask to track matched elements in group 2
        memset(dp, -1, sizeof(dp));  
        
        return solve(0, 0, matchedMask, cost);
    }
};

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