Introduction

In this blog post, we will explore how to solve the LeetCode problem "1595" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two groups of points where the first group has size1 points, the second group has size2 points, and size1 >= size2. The cost of the connection between any two points are given in an size1 x size2 matrix where cost[i][j] is the cost of connecting point i of the first group and point j of the second group. 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. Return the minimum cost it takes to connect the two groups. Example 1: Input: cost = [[15, 96], [36, 2]] Output: 17 Explanation: The optimal way of connecting the groups is: 1--A 2--B This results in a total cost of 17. Example 2: Input: cost = [[1, 3, 5], [4, 1, 1], [1, 5, 3]] Output: 4 Explanation: 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. Example 3: Input: cost = [[2, 5, 1], [3, 4, 7], [8, 1, 2], [6, 2, 4], [3, 8, 8]] Output: 10 Constraints: size1 == cost.length size2 == cost[i].length 1 <= size1, size2 <= 12 size1 >= size2 0 <= cost[i][j] <= 100

Explanation

Here's a breakdown of the solution:

• Dynamic Programming with Bitmasking: Represent the connections of the second group's points using a bitmask. The i-th bit being set means the i-th point in the second group is connected. We use DP to find the minimum cost to connect the first group's points such that all points in the second group are connected.

• State Definition: dp[i][mask] stores the minimum cost to connect the first i points (0 to i-1) in the first group such that the points represented by the mask in the second group are connected.

• Transitions: For each point i in the first group, we have two choices: either connect it to one or more points in the second group, or skip it (implicitly handled by taking the min of previous states). If connecting it, we iterate through all possible subsets of the second group and calculate the cost.

• Runtime Complexity: O(size1 * 3^size2).

• Storage Complexity: O(size1 * 2^size2).

Code

    def connect_two_groups(cost):
    size1 = len(cost)
    size2 = len(cost[0])
    dp = {}

    def solve(i, mask):
        if i == size1:
            if mask == (1 << size2) - 1:
                return 0
            else:
                return float('inf')

        if (i, mask) in dp:
            return dp[(i, mask)]

        ans = float('inf')

        # Option 1: Don't connect i-th point to any point in group 2
        ans = min(ans, solve(i + 1, mask))

        # Option 2: Connect i-th point to one or more points in group 2
        for submask in range(1 << size2):
            current_cost = 0
            new_mask = mask
            if submask > 0:
                for j in range(size2):
                    if (submask >> j) & 1:
                        current_cost += cost[i][j]
                        new_mask |= (1 << j)
                ans = min(ans, current_cost + solve(i + 1, new_mask))

        dp[(i, mask)] = ans
        return ans

    return solve(0, 0)