1584. Min Cost to Connect All Points
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Difficulty: medium Acceptance: 69% Topics: Array, Union Find, Graph, Minimum Spanning Tree
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Reference solution (spoiler · python)
# Time: O(n^2)
# Space: O(n)
class Solution(object):
def minCostConnectPoints(self, points):
"""
:type points: List[List[int]]
:rtype: int
"""
result, u = 0, 0 # we can start from any node as u
dist = [float("inf")]*len(points)
lookup = set()
for _ in xrange(len(points)-1):
x0, y0 = points[u]
lookup.add(u)
for v, (x, y) in enumerate(points):
if v in lookup:
continue
dist[v] = min(dist[v], abs(x-x0) + abs(y-y0))
val, u = min((val, v) for v, val in enumerate(dist))
dist[u] = float("inf") # used
result += val
return result
# Time: O(eloge) = O(n^2 * logn)
# Space: O(e) = O(n^2)
# kruskal's algorithm
class UnionFind(object): # Time: O(n * α(n)), Space: O(n)
def __init__(self, n):
self.set = range(n)
self.rank = [0]*n
def find_set(self, x):
stk = []
while self.set[x] != x: # path compression
stk.append(x)
x = self.set[x]
while stk:
self.set[stk.pop()] = x
return x
def union_set(self, x, y):
x_root, y_root = map(self.find_set, (x, y))
if x_root == y_root:
return False
if self.rank[x_root] < self.rank[y_root]: # union by rank
self.set[x_root] = y_root
elif self.rank[x_root] > self.rank[y_root]:
self.set[y_root] = x_root
else:
self.set[y_root] = x_root
self.rank[x_root] += 1
return True
class Solution2(object):
def minCostConnectPoints(self, points):
"""
:type points: List[List[int]]
:rtype: int
"""
edges = []
for u in xrange(len(points)):
for v in xrange(u+1, len(points)):
edges.append((u, v, abs(points[v][0]-points[u][0]) + abs(points[v][1]-points[u][1])))
edges.sort(key=lambda x: x[2])
result = 0
union_find = UnionFind(len(points))
for u, v, val in edges:
if union_find.union_set(u, v):
result += val
return result
Solution from kamyu104/LeetCode-Solutions · MIT
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