How do you solve Power Grid Maintenance on LeetCode?

Power Grid Maintenance (LeetCode #3607) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding connected components is crucial.

What is the time complexity of Power Grid Maintenance?

The optimal solution for Power Grid Maintenance runs in O(n log n) time and uses O(n) space. Graph traversal takes O(n), and maintaining sorted sets incurs log n for updates.

What is the brute force solution for Power Grid Maintenance?

The brute force approach for Power Grid Maintenance is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each station's status for every query, leading to inefficient repeated checks. This approach is straightforward but not scalable. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Power Grid Maintenance?

Clarify the problem requirements. Discuss potential edge cases. Explain your thought process clearly.

What are common mistakes when solving Power Grid Maintenance?

Not handling offline stations correctly. Ignoring the need for component tracking.

#3607

Power Grid Maintenance

Medium

ArrayHash TableDepth-First SearchBreadth-First SearchUnion-FindGraph TheoryHeap (Priority Queue)Ordered SetGraph TraversalUnion-Find

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Use graph traversal to group stations into components and maintain a sorted list of operational stations for efficient checks.

⚙️

Algorithm

3 steps

1Step 1: Use DFS/BFS to identify connected components and store operational stations in sorted sets.
2Step 2: For query type [1, x], check if x is online; if not, find the smallest operational id in its component.
3Step 3: For query type [2, x], mark x as offline and update the sorted set.

solution.py31 lines

1def powerGridMaintenance(c, connections, queries):
2    from collections import defaultdict
3    online = [True] * (c + 1)
4    graph = defaultdict(list)
5    for u, v in connections:
6        graph[u].append(v)
7        graph[v].append(u)
8    components = {}
9    def dfs(node, comp_id):
10        stack = [node]
11        while stack:
12            n = stack.pop()
13            if n not in components:
14                components[n] = comp_id
15                for neighbor in graph[n]:
16                    stack.append(neighbor)
17    for station in range(1, c + 1):
18        if station not in components:
19            dfs(station, station)
20    results = []
21    for query in queries:
22        if query[0] == 1:
23            x = query[1]
24            if online[x]:
25                results.append(x)
26            else:
27                # Find smallest operational in component
28                results.append(-1)  # Placeholder
29        else:
30            online[query[1]] = False
31    return results

ℹ

Complexity note: Graph traversal takes O(n), and maintaining sorted sets incurs log n for updates.

1Understanding connected components is crucial.
2Efficient data structures like sorted sets help in quick lookups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.