<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>

from typing import List

class Solution:
    def connectTwoGroups(self, cost: List[List[int]]) -> int:
        num_group1 = len(cost)
        num_group2 = len(cost[0])
        memo = [[None] * (1 << num_group2) for _ in range(num_group1)]
        
        # minCostPerPoint[j] := the minimum cost of connecting group2's point j
        minCostPerPoint = [float('inf')] * num_group2
        
        for j in range(num_group2):
            minCostIndex = 0
            for i in range(1, num_group1):
                if cost[i][j] < cost[minCostIndex][j]:
                    minCostIndex = i
            minCostPerPoint[j] = cost[minCostIndex][j]

        return self._calculateMinimumCost(cost, 0, 0, minCostPerPoint, memo)

    def _calculateMinimumCost(self, cost: List[List[int]], index: int, bitmask: int, minCostPerPoint: List[int], memo: List[List[int]]) -> int:
        if index == len(cost):
            # All points in group 1 are connected, so greedily assign the minimum cost for the unconnected points of group2.
            total_cost = 0
            for j in range(len(cost[0])):
                if not (bitmask >> j & 1):
                    total_cost += minCostPerPoint[j]
            return total_cost
        if memo[index][bitmask] is not None:
            return memo[index][bitmask]

        min_cost = float('inf')
        for j in range(len(cost[0])):
            min_cost = min(min_cost, cost[index][j] + self._calculateMinimumCost(cost, index + 1, bitmask | (1 << j), minCostPerPoint, memo))
        memo[index][bitmask] = min_cost
        return min_cost

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