# Minimum Cost To Connect Two Groups Of Points

#### You are given two groups of points where the first group has ‘N’ points, the second group has ‘M’ points, and ‘N’ >= ‘M’.

#### You are also given a “cost” matrix of ‘N’ * ‘M’ dimensions whose (i, j)th element denotes the cost of the connection between ith point of group 1 and jth point of group 2. The groups are connected if each point in both groups is connected to one or more points in the opposite group. 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.

#### Now you are supposed to find the minimum cost it takes to connect the two groups.

##### Input Format :

```
The first line contains an integer ‘T’ denoting the number of test cases. Then each test case follows.
The first input line of each test case contains two integers ‘N’ and ‘M’, denoting the number of rows and columns of the “cost” matrix.
Each of the next ‘N’ lines contains ‘M’ space-separated integers denoting the elements of the “cost” matrix.
```

##### Output Format :

```
For each test case, print an integer denoting the minimum cost it takes to connect the two groups.
Print the output of each test case in a separate line.
```

##### Note :

```
You are not required to print the expected output; it has already been taken care of. Just implement the function.
```

##### Constraints :

```
1 <= T <= 5
1 <= N <= 10
1 <= M <= 10
0 <= cost[i][j] <= 100
Time limit: 1 sec
```

Let us try to break down the problem into sub-problems. Consider a function

**getMinCost(int id, int mask)**

which returns the minimum cost to connect the first “id” elements of group 1 to a subset of group 2 which is denoted by “mask”. Here, the “mask” is an integer whose jth bit (from Least significant bit) is set if the jth element (0th based indexing) of group 2 is already included in the subset.

Now, let us see what happens when we try to connect the id-th element of group 1 with some (one or more) elements of group 2. We’ll have to add the cost associated with the new connection and also we’ll have to update the “mask” if a new element from group 2 is included.

Now, consider the following steps for implementing the function getMinCost():

- Base Case:
- Check if “id” is equal to -1.
- If “mask” is equal to (1 << M) - 1, which means every point of the second group is included in the connection. So, return 0.
- Otherwise, return a very large integer.

- Check if “id” is equal to -1.
- Initialise a “ans” variable with a very large integer which will store the result of the current function call.
- Now, iterate through the points of the second group using the variable ‘j’. These are the following two options possible.
- id-th element is connected to only one point from 2nd group. For this case: ans = min(ans,cost[id][j] + getMinCost(id - 1, mask | (1 << j))
- Id-th element is connected to more than one points from group 2 and j-th point is one of them. For this case, if jth bit is not set:
**ans = min(ans, cost[id][j] + getMinCost(id, mask | (1 << j))**

- Return “ans”.

Let us observe the recursion tree for first test case of sample test 1 where

“cost” = [[ 1, 2], [2, 3], [4, 1]].

After observing the tree, we’ll find that there are some redundant function calls which means that there are some overlapping sub-problems. The repetition of such sub-problems suggests that we can use dynamic programming to optimise our approach.

The key idea behind a dynamic programming approach is to use memoisation, i.e. we’ll save the result of our sub-problem in a matrix so that it can be used later on.

Create a dynamic programming matrix “dp” of ‘N’ * (2 ^ ‘M’) dimension which will be used to store the results to avoid redundant function calls. dp[id[mask] will store the minimum cost to connect the first “id” elements of group 1 to a subset of group 2 which is denoted by “mask”. Here, the “mask” is an integer whose jth bit (from the Least significant bit) is set if the jth element (0th based indexing) of group 2 is already included in the subset.

Now, consider the following steps for implementing the function getMinCost():

- Base Case:
- Check if “id” is equal to -1.
- If “mask” is equal to (1 << M) - 1, which means every point of the second group is included in the connection. So, return 0.
- Otherwise, return a very large integer.

- Check if “id” is equal to -1.
- Check if the answer for this problem already exists in the “dp” matrix. If it exists, return the answer right away.
- Initialise a “ans” variable with a very large integer which will store the result of the current function call.
- Now, iterate through the points of the second group using the variable ‘j’. These are the following two options possible.
- id-th element is connected to only one point from 2nd group. For this case:
**ans = min(ans,cost[id][j] + getMinCost(id - 1, mask | (1 << j))** - Id-th element is connected to more than one points from group 2 and j-th point is one of them. For this case, if jth bit is not set:
**ans = min(ans, cost[id][j] + getMinCost(id , mask | (1 << j))**

- id-th element is connected to only one point from 2nd group. For this case:
- Store the “ans” in “dp[id][mask]” matrix to avoid redundant function calls and also return it.