How do you solve Reachable Nodes In Subdivided Graph on LeetCode?

Reachable Nodes In Subdivided Graph (LeetCode #882) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(E log V) time complexity and O(V) space complexity. Key insight: Understanding graph traversal is crucial.

What is the time complexity of Reachable Nodes In Subdivided Graph?

The optimal solution for Reachable Nodes In Subdivided Graph runs in O(E log V) time and uses O(V) space. The complexity is O(E log V) due to the priority queue operations, where E is the number of edges and V is the number of vertices.

What is the brute force solution for Reachable Nodes In Subdivided Graph?

The brute force approach for Reachable Nodes In Subdivided Graph is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves constructing the entire subdivided graph and then performing a depth-first search (DFS) or breadth-first search (BFS) to count reachable nodes from node 0. This method is straightforward but inefficient due to the large number of nodes created. The optimal Optimal Solution approach improves this to O(E log V).

What algorithm or data structure is used to solve Reachable Nodes In Subdivided Graph?

Reachable Nodes In Subdivided Graph uses the following concepts: Graph Theory, Heap (Priority Queue), Shortest Path. The recommended patterns to study are: Graph Traversal, Priority Queue, Dijkstra's Algorithm.

What are the interview tips for Reachable Nodes In Subdivided Graph?

Clarify the problem and constraints before coding. Discuss your thought process and approach with the interviewer. Test edge cases to ensure your solution is robust.

What are common mistakes when solving Reachable Nodes In Subdivided Graph?

Not considering the distance when adding nodes to the heap. Failing to mark nodes as visited, leading to infinite loops.

#882

Reachable Nodes In Subdivided Graph

Hard

Graph TheoryHeap (Priority Queue)Shortest PathGraph TraversalPriority QueueDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(E log V)Space O(V)

The optimal solution uses a priority queue (min-heap) to efficiently explore the graph, prioritizing nodes based on the distance traveled. This allows us to explore the most promising paths first and ensures we stay within the maxMoves limit while counting reachable nodes.

⚙️

Algorithm

3 steps

1Step 1: Create a graph representation using an adjacency list.
2Step 2: Use a min-heap to explore nodes starting from node 0, keeping track of the distance traveled.
3Step 3: For each node, calculate the maximum number of new nodes that can be reached from its edges and update the reachable count accordingly.

solution.py26 lines

1# Full working Python code
2import heapq
3from collections import defaultdict
4
5def reachableNodes(edges, maxMoves, n):
6    graph = defaultdict(list)
7    for u, v, cnt in edges:
8        graph[u].append((v, cnt))
9        graph[v].append((u, cnt))
10
11    heap = [(0, 0)]  # (distance, node)
12    reachable = 0
13    visited = set()
14    while heap:
15        dist, node = heapq.heappop(heap)
16        if node in visited:
17            continue
18        visited.add(node)
19        reachable += 1
20        for neighbor, cnt in graph[node]:
21            if neighbor not in visited:
22                new_dist = dist + cnt + 1
23                if new_dist <= maxMoves:
24                    heapq.heappush(heap, (new_dist, neighbor))
25
26    return reachable

ℹ

Complexity note: The complexity is O(E log V) due to the priority queue operations, where E is the number of edges and V is the number of vertices.

1Understanding graph traversal is crucial.
2Using a priority queue can optimize the search process.

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