<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:
    def connectTwoGroups(self, cost: List[List[int]]) -> int:
        from functools import lru_cache

        size1, size2 = len(cost), len(cost[0])
        all_mask = (1 << size2) - 1

        # Utility to calculate min cost from each point in group 2 to group 1
        min_cost_to_group1 = [min(cost[i][j] for i in range(size1)) for j in range(size2)]

        @lru_cache(None)
        def dp(idx: int, mask: int) -> int:
            # Base case: if all points in the first group are considered
            if idx == size1:
                # Cost to connect remaining unconnected points in group 2
                return sum(min_cost_to_group1[j] for j in range(size2) if not mask & (1 << j))

            # Try connecting current point of group 1 to each point in group 2
            min_total_cost = float('inf')
            for j in range(size2):
                min_total_cost = min(min_total_cost, cost[idx][j] + dp(idx + 1, mask | (1 << j)))

            return min_total_cost

        return dp(0, 0)

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