Leetcode 1584. Min Cost to Connect All Points
You are given an array points representing integer coordinates of some points on a 2D-plane, where points = .The cost of connecting two points and is the manhattan distance between them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.
Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.
Example 1:
Input: points = [,,,,]
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.
Example 2:
Input: points = [,[-2,5],[-4,1]]
Output: 18
Example 3:
Input: points = [,,,[-1,1]]
Output: 4
Example 4:
Input: points = [[-1000000,-1000000],]
Output: 4000000
Example 5:
Input: points = []
Output: 0
Constraints:
1 <= points.length <= 1000
-106 <= xi, yi <= 106
All pairs (xi, yi) are distinct.
class UnionFind:
def __init__(self, n: int):
self.parent = list(range(n))
self.size = * n
def find(self, x):
if self.parent != x:
self.parent = self.find(self.parent)
return self.parent
def union(self, x, y):
rx, ry = self.find(x), self.find(y)
if rx == ry: return False
if self.size < self.size:
rx, ry = ry, rx
self.parent = rx
self.size += self.size
return True
class Solution:
def minCostConnectPoints(self, points: List]) -> int:
N = len(points)
edges = []
res = 0
for i in range(N):
for j in range(i + 1, N):
distance = abs(points - points)
distance += abs(points - points)
edges.append((distance, i, j))
edges.sort()
unionfind = UnionFind(N)
for edge in edges:
dis, i, j = edge
if unionfind.union(i, j):
res += dis
return res
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