Leetcode #1595: Minimum Cost to Connect Two Groups of Points
In this guide, we solve Leetcode #1595 Minimum Cost to Connect Two Groups of Points in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Dynamic Programming, Bitmask, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def connectTwoGroups(self, cost: List[List[int]]) -> int:
m, n = len(cost), len(cost[0])
f = [[inf] * (1 << n) for _ in range(m + 1)]
f[0][0] = 0
for i in range(1, m + 1):
for j in range(1 << n):
for k in range(n):
if (j >> k & 1) == 0:
continue
c = cost[i - 1][k]
x = min(f[i][j ^ (1 << k)], f[i - 1][j], f[i - 1][j ^ (1 << k)]) + c
f[i][j] = min(f[i][j], x)
return f[m][-1]
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.