How do you solve Network Delay Time on LeetCode?

Network Delay Time (LeetCode #743) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O((E + V) log V) time complexity and O(V) space complexity. Key insight: Understanding graph traversal is crucial for solving shortest path problems.

What is the brute force solution for Network Delay Time?

The brute force approach for Network Delay Time is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach attempts to simulate the signal propagation by checking all possible paths from the starting node to each other node. It calculates the time taken for each path and keeps track of the maximum time needed for all nodes to receive the signal. The optimal Optimal Solution approach improves this to O((E + V) log V).

What are the interview tips for Network Delay Time?

Clearly explain your thought process and the choice of algorithm. Discuss edge cases, such as disconnected nodes or cycles. Practice implementing Dijkstra's algorithm, as it's a common pattern in graph problems.

What are common mistakes when solving Network Delay Time?

Failing to handle unreachable nodes correctly, often returning incorrect results. Not using a priority queue or similar structure, leading to inefficient solutions.

#743

Network Delay Time

Medium

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

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal solution uses Dijkstra's algorithm, which efficiently finds the shortest paths from the starting node to all other nodes in a graph with non-negative weights. This approach is faster than brute force because it prioritizes exploring the shortest paths first.

⚙️

Algorithm

5 steps

1Step 1: Create a priority queue to store nodes along with their current travel time.
2Step 2: Initialize the queue with the starting node and its time (0).
3Step 3: While the queue is not empty, extract the node with the smallest time, mark it as visited, and update the times for its neighbors.
4Step 4: If a neighbor can be reached in a shorter time, update its time and add it to the queue.
5Step 5: After processing all nodes, return the maximum time from the starting node to all other nodes, or -1 if any node is unreachable.

solution.py22 lines

1# Full working Python code
2import heapq
3from collections import defaultdict
4
5def networkDelayTime(times, n, k):
6    graph = defaultdict(list)
7    for u, v, w in times:
8        graph[u].append((v, w))
9    min_time = [float('inf')] * (n + 1)
10    min_time[k] = 0
11    pq = [(0, k)]  # (time, node)
12    while pq:
13        curr_time, curr_node = heapq.heappop(pq)
14        if curr_time > min_time[curr_node]:
15            continue
16        for neighbor, weight in graph[curr_node]:
17            time = curr_time + weight
18            if time < min_time[neighbor]:
19                min_time[neighbor] = time
20                heapq.heappush(pq, (time, neighbor))
21    result = max(min_time[1:])
22    return result if result < float('inf') else -1

ℹ

Complexity note: This complexity arises because we process each edge and node, and the priority queue operations take logarithmic time relative to the number of nodes.

1Understanding graph traversal is crucial for solving shortest path problems.
2Using a priority queue helps efficiently manage the exploration of nodes based on the shortest known distance.

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