How do you solve Minimum Cost to Connect Two Groups of Points on LeetCode?

Minimum Cost to Connect Two Groups of Points (LeetCode #1595) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^m * m) time complexity and O(2^m) space complexity. Key insight: Using bitmasking allows us to efficiently track which points are connected.

What is the brute force solution for Minimum Cost to Connect Two Groups of Points?

The brute force approach for Minimum Cost to Connect Two Groups of Points is Brute Force, with O(n² * 2^n) time complexity and O(1) space complexity. The brute force approach tries every possible combination of connections between the two groups to find the minimum cost. While it's straightforward, it quickly becomes impractical as the number of points increases. The optimal Optimal Solution approach improves this to O(n * 2^m * m).

What are the interview tips for Minimum Cost to Connect Two Groups of Points?

Explain your thought process clearly when discussing the DP approach. Be prepared to discuss time and space complexity in detail. Practice similar problems to become comfortable with bitmasking and DP.

What are common mistakes when solving Minimum Cost to Connect Two Groups of Points?

Forgetting to initialize the DP array properly. Not considering all possible connections when updating the DP state.

#1595

Minimum Cost to Connect Two Groups of Points

Hard

ArrayDynamic ProgrammingBit ManipulationMatrixBitmaskDynamic ProgrammingBitmasking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * 2^n)	O(n * 2^m * m)
Space	O(1)	O(2^m)

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Intuition

Time O(n * 2^m * m)Space O(2^m)

The optimal solution uses dynamic programming with bitmasking to efficiently compute the minimum cost. This approach reduces the number of combinations we need to check by storing intermediate results.

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Algorithm

3 steps

1Step 1: Initialize a DP array where dp[mask] represents the minimum cost to connect points in the first group represented by 'mask'.
2Step 2: Iterate through all possible masks and for each mask, calculate the cost to connect additional points from the second group.
3Step 3: Update the DP array with the minimum costs found.

solution.py9 lines

1def minCost(cost):
2    n, m = len(cost), len(cost[0])
3    dp = [float('inf')] * (1 << m)
4    dp[0] = 0
5    for i in range(n):
6        for mask in range((1 << m) - 1, -1, -1):
7            for j in range(m):
8                dp[mask | (1 << j)] = min(dp[mask | (1 << j)], dp[mask] + cost[i][j])
9    return min(dp[(1 << m) - 1], float('inf'))

ℹ

Complexity note: This complexity arises because we iterate through all points in the first group (n), and for each point, we check all possible connections to the second group (2^m). This is significantly more efficient than the brute force approach.

1Using bitmasking allows us to efficiently track which points are connected.
2Dynamic programming helps us avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.