3863. Power Grid Maintenance

Array Hash Table Depth-First Search Breadth-First Search Union Find Graph Heap (Priority Queue) Ordered Set

Problem - Power Grid Maintenance

Medium

You are given an integer c representing c power stations, each with a unique identifier id from 1 to c (1‑based indexing).

These stations are interconnected via n bidirectional cables, represented by a 2D array connections, where each element connections[i] = [ui, vi] indicates a connection between station ui and station vi. Stations that are directly or indirectly connected form a power grid.

Initially, all stations are online (operational).

You are also given a 2D array queries, where each query is one of the following two types:

• [1, x]: A maintenance check is requested for station x. If station x is online, it resolves the check by itself. If station x is offline, the check is resolved by the operational station with the smallest id in the same power grid as x. If no operational station exists in that grid, return -1.

• [2, x]: Station x goes offline (i.e., it becomes non-operational).

Return an array of integers representing the results of each query of type [1, x] in the order they appear.

Note: The power grid preserves its structure; an offline (non‑operational) node remains part of its grid and taking it offline does not alter connectivity.

 

Example 1:

Input: c = 5, connections = [[1,2],[2,3],[3,4],[4,5]], queries = [[1,3],[2,1],[1,1],[2,2],[1,2]]

Output: [3,2,3]

Explanation:

• Initially, all stations {1, 2, 3, 4, 5} are online and form a single power grid.
• Query [1,3]: Station 3 is online, so the maintenance check is resolved by station 3.
• Query [2,1]: Station 1 goes offline. The remaining online stations are {2, 3, 4, 5}.
• Query [1,1]: Station 1 is offline, so the check is resolved by the operational station with the smallest id among {2, 3, 4, 5}, which is station 2.
• Query [2,2]: Station 2 goes offline. The remaining online stations are {3, 4, 5}.
• Query [1,2]: Station 2 is offline, so the check is resolved by the operational station with the smallest id among {3, 4, 5}, which is station 3.

Example 2:

Input: c = 3, connections = [], queries = [[1,1],[2,1],[1,1]]

Output: [1,-1]

Explanation:

• There are no connections, so each station is its own isolated grid.
• Query [1,1]: Station 1 is online in its isolated grid, so the maintenance check is resolved by station 1.
• Query [2,1]: Station 1 goes offline.
• Query [1,1]: Station 1 is offline and there are no other stations in its grid, so the result is -1.

 

Constraints:

• 1 <= c <= 105
• 0 <= n == connections.length <= min(105, c * (c - 1) / 2)
• connections[i].length == 2
• 1 <= ui, vi <= c
• ui != vi
• 1 <= queries.length <= 2 * 105
• queries[i].length == 2
• queries[i][0] is either 1 or 2.
• 1 <= queries[i][1] <= c

Solutions

Python3

class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, val):
        if self.p[val] != val:
            self.p[val] = self.find(self.p[val])
        return self.p[val]

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True


class Solution:
    def processQueries(self, c: int, connections: List[List[int]], queries: List[List[int]]) -> List[int]:
        union_find = UnionFind(c + 1)
        for u, v in connections:
            union_find.union(u, v)

        sorted_list = [SortedList() for _ in range(c + 1)]
        for i in range(1, c + 1):
            sorted_list[union_find.find(i)].add(i)

        result = []
        for a, x in queries:
            root = union_find.find(x)
            if a == 1:
                if x in sorted_list[root]:
                    result.append(x)
                elif len(sorted_list[root]):
                    result.append(sorted_list[root][0])
                else:
                    result.append(-1)
            else:
                sorted_list[root].discard(x)

        return result

Submission Stats:

• Runtime: 1478 ms (6.86%)
• Memory: 122.4 MB (13.48%)