LeetCode #3607 — MEDIUM

Power Grid Maintenance

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

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
Patterns Used
🔗Union-Find→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Union-Find · Segment Tree

Example 1

5
[[1,2],[2,3],[3,4],[4,5]]
[[1,3],[2,1],[1,1],[2,2],[1,2]]

Example 2

3
[]
[[1,1],[2,1],[1,1]]
Step 02

Core Insight

What unlocks the optimal approach

  • Use DFS or BFS to assign each station a component ID
  • For each component, maintain a sorted set of online station IDs
  • For query <code>[2, x]</code>, remove <code>x</code> from the set of its component
  • For query <code>[1, x]</code>, if <code>x</code> is in its component’s set return <code>x</code>; otherwise if the set is non-empty return its smallest element; else return <code>-1</code>
  • Precompute all components and then handle each query in O(log n) time using the sorted sets
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3607: Power Grid Maintenance
class UnionFind {
    private final int[] p;
    private final int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    public int[] processQueries(int c, int[][] connections, int[][] queries) {
        UnionFind uf = new UnionFind(c + 1);
        for (int[] e : connections) {
            uf.union(e[0], e[1]);
        }

        TreeSet<Integer>[] st = new TreeSet[c + 1];
        Arrays.setAll(st, k -> new TreeSet<>());
        for (int i = 1; i <= c; i++) {
            int root = uf.find(i);
            st[root].add(i);
        }

        List<Integer> ans = new ArrayList<>();
        for (int[] q : queries) {
            int a = q[0], x = q[1];
            int root = uf.find(x);

            if (a == 1) {
                if (st[root].contains(x)) {
                    ans.add(x);
                } else if (!st[root].isEmpty()) {
                    ans.add(st[root].first());
                } else {
                    ans.add(-1);
                }
            } else {
                st[root].remove(x);
            }
        }

        return ans.stream().mapToInt(Integer::intValue).toArray();
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O((c + n + q)
Space
O(c)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND
O(α(n)) time
O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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