Time: 2386 ms (5.39%), Space: 234.3 MB (5.15%) - LeetHub

Author: hogan-tech

3863-power-grid-maintenance/3863-power-grid-maintenance.py

@@ -0,0 +1,68 @@
+# time complexity: O((c + q) * α(c))
+# space complexity: O(c)
+from typing import List
+from collections import defaultdict
+from sortedcontainers import SortedSet
+
+class Solution:
+    def processQueries(self, c: int, connections: List[List[int]], queries: List[List[int]]) -> List[int]:
+        parent = list(range(c + 1))
+
+        def find(x):
+            while x != parent[x]:
+                parent[x] = parent[parent[x]]
+                x = parent[x]
+            return x
+
+        def union(x, y):
+            px, py = find(x), find(y)
+            if px != py:
+                parent[py] = px
+
+        for u, v in connections:
+            union(u, v)
+
+        groupMembers = defaultdict(list)
+        for node in range(1, c + 1):
+            root = find(node)
+            groupMembers[root].append(node)
+
+        componentMap = dict()
+        for root, members in groupMembers.items():
+            componentMap[root] = SortedSet(members)
+
+        nodeToRoot = {node: find(node) for node in range(1, c + 1)}
+
+        online = [True] * (c + 1)
+
+        result = []
+
+        for type, x in queries:
+            if type == 1:  
+                if online[x]:
+                    result.append(x)
+                else:
+                    root = nodeToRoot[x]
+                    candidates = componentMap[root]
+                    if candidates:
+                        result.append(candidates[0])
+                    else:
+                        result.append(-1)
+            else:  
+                if online[x]:
+                    online[x] = False
+                    root = nodeToRoot[x]
+                    componentMap[root].discard(x)
+
+        return result
+
+
+
+c = 5
+connections = [[1, 2], [2, 3], [3, 4], [4, 5]]
+queries = [[1, 3], [2, 1], [1, 1], [2, 2], [1, 2]]
+print(Solution().processQueries(c, connections, queries))
+c = 3
+connections = []
+queries = [[1, 1], [2, 1], [1, 1]]
+print(Solution().processQueries(c, connections, queries))