3607. Power Grid Maintenance

Problem Link

Description

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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

Solution

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Python3

class DSU:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0 for _ in range(n)]
 
    def find(self, x):
        if self.parent[x] == x:
            return self.parent[x]
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
 
    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)
 
        if (pu == pv): return
 
        if self.rank[pu] < self.rank[pv]:
            pu, pv = pv, pu
 
        # ensure self.rank[pu] >= self.rank[pv]
        self.parent[pv] = pu
        if self.rank[pu] == self.rank[pv]:
            self.rank[pu] += 1
 
class Solution:
    def processQueries(self, N: int, connections: List[List[int]], queries: List[List[int]]) -> List[int]:
        uf = DSU(N + 1)
        disconnected = set()
 
        for a, b in connections:
            uf.union(a, b)
 
        grouped = defaultdict(list)
        for node in range(1, N + 1):
            heappush(grouped[uf.find(node)], node)
 
        res = []
        for t, node in queries:
            if t == 1:
                if node not in disconnected:
                    res.append(node)
                    continue
                
                A = grouped[uf.find(node)]
                while A and A[0] in disconnected:
                    heappop(A)
 
                if not A:
                    res.append(-1)
                else:
                    res.append(A[0])
            else:
                disconnected.add(node)
 
        return res