<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 connectTwoGroups(vector<vector<int>>& cost) {
        int size1 = cost.size();
        int size2 = cost[0].size();
        
        // The maximum possible mask for group 2 (since size2 <= 12, the mask can go up to 2^12 - 1).
        int full_mask = (1 << size2) - 1;
        
        // DP array initialized to a large value (infinity).
        vector<vector<int>> dp(size1 + 1, vector<int>(1 << size2, INT_MAX));
        
        // Base case: No points in group 1 are connected, so no cost.
        dp[0][0] = 0;
        
        // DP transition
        for (int i = 0; i < size1; ++i) {
            for (int mask = 0; mask <= full_mask; ++mask) {
                if (dp[i][mask] == INT_MAX) continue;
                
                // Try connecting group 1 point `i` to group 2 point `j`.
                for (int j = 0; j < size2; ++j) {
                    int new_mask = mask | (1 << j);  // Add j-th point of group 2 to the mask
                    dp[i + 1][new_mask] = min(dp[i + 1][new_mask], dp[i][mask] + cost[i][j]);
                    
                    // Also consider the case where we don't move to the next point in group 1 yet
                    dp[i][new_mask] = min(dp[i][new_mask], dp[i][mask] + cost[i][j]);
                }
            }
        }
        
        // The answer is the minimum cost to connect all points in group 2 (full_mask) when all points in group 1 are connected.
        return dp[size1][full_mask];
    }
};
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