Leetcode #1584: Min Cost to Connect All Points
In this guide, we solve Leetcode #1584 Min Cost to Connect All 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 an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi]. The cost of connecting two points [xi, yi] and [xj, yj] is the manhattan distance between them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Union Find, Graph, Array, Minimum Spanning Tree
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
Output: 20
Explanation:
We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.
Python Solution
class Solution:
def minCostConnectPoints(self, points: List[List[int]]) -> int:
n = len(points)
g = [[0] * n for _ in range(n)]
dist = [inf] * n
vis = [False] * n
for i, (x1, y1) in enumerate(points):
for j in range(i + 1, n):
x2, y2 = points[j]
t = abs(x1 - x2) + abs(y1 - y2)
g[i][j] = g[j][i] = t
dist[0] = 0
ans = 0
for _ in range(n):
i = -1
for j in range(n):
if not vis[j] and (i == -1 or dist[j] < dist[i]):
i = j
vis[i] = True
ans += dist[i]
for j in range(n):
if not vis[j]:
dist[j] = min(dist[j], g[i][j])
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.