How do you solve Find Edges in Shortest Paths on LeetCode?

Find Edges in Shortest Paths (LeetCode #3123) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O((n + m) log n) time complexity and O(n) space complexity. Key insight: Using Dijkstra's algorithm allows us to efficiently find shortest paths in weighted graphs.

What is the time complexity of Find Edges in Shortest Paths?

The optimal solution for Find Edges in Shortest Paths runs in O((n + m) log n) time and uses O(n) space. Dijkstra's algorithm runs in O((n + m) log n) time due to the priority queue operations, making it efficient for sparse graphs.

What is the brute force solution for Find Edges in Shortest Paths?

The brute force approach for Find Edges in Shortest Paths is Brute Force, with O(n²) time complexity and O(1) space complexity. We can find all shortest paths from node 0 to node n-1 by exploring every possible path in the graph. This approach is simple but inefficient, as it may involve checking many paths. The optimal Optimal Solution approach improves this to O((n + m) log n).

What are the interview tips for Find Edges in Shortest Paths?

Always clarify the problem and constraints before diving into coding. Discuss your thought process and potential approaches with the interviewer. Optimize your solution step-by-step, explaining why each optimization is necessary.

What are common mistakes when solving Find Edges in Shortest Paths?

Not considering both directions when checking if an edge is part of a shortest path. Overcomplicating the brute force approach instead of focusing on systematic exploration.

#3123

Find Edges in Shortest Paths

Hard

Depth-First SearchBreadth-First SearchGraph TheoryHeap (Priority Queue)Shortest PathDijkstra's AlgorithmGraph TraversalPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O((n + m) log n)Space O(n)

We can use Dijkstra's algorithm to find the shortest path from node 0 to all other nodes and from node n-1 to all other nodes. This allows us to efficiently determine if an edge is part of any shortest path.

⚙️

Algorithm

3 steps

1Step 1: Use Dijkstra's algorithm to find the shortest distance from node 0 to all other nodes.
2Step 2: Use Dijkstra's algorithm again to find the shortest distance from node n-1 to all other nodes.
3Step 3: For each edge, check if the sum of the distances from node 0 to the start of the edge, the weight of the edge, and the distance from the end of the edge to node n-1 equals the shortest distance from node 0 to n-1.

solution.py32 lines

1import heapq
2from collections import defaultdict
3
4def find_edges_in_shortest_paths(n, edges):
5    graph = defaultdict(list)
6    for a, b, w in edges:
7        graph[a].append((b, w))
8        graph[b].append((a, w))
9
10    def dijkstra(start):
11        distances = [float('inf')] * n
12        distances[start] = 0
13        pq = [(0, start)]
14        while pq:
15            curr_dist, node = heapq.heappop(pq)
16            if curr_dist > distances[node]: continue
17            for neighbor, weight in graph[node]:
18                distance = curr_dist + weight
19                if distance < distances[neighbor]:
20                    distances[neighbor] = distance
21                    heapq.heappush(pq, (distance, neighbor))
22        return distances
23
24    dist_from_start = dijkstra(0)
25    dist_from_end = dijkstra(n - 1)
26    min_distance = dist_from_start[n - 1]
27    answer = [False] * len(edges)
28    for i, (a, b, w) in enumerate(edges):
29        if dist_from_start[a] + w + dist_from_end[b] == min_distance or \
30           dist_from_start[b] + w + dist_from_end[a] == min_distance:
31            answer[i] = True
32    return answer

ℹ

Complexity note: Dijkstra's algorithm runs in O((n + m) log n) time due to the priority queue operations, making it efficient for sparse graphs.

1Using Dijkstra's algorithm allows us to efficiently find shortest paths in weighted graphs.
2Checking both directions (from start and end) helps in determining if an edge is part of the shortest path.

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