LeetCode 743: Network Delay Time Solution

Master LeetCode problem 743 (Network Delay Time), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

743. Network Delay Time

Medium

Depth-First Search Breadth-First Search Graph Heap (Priority Queue)Shortest Path

Problem Explanation

Explanation

To solve this problem, we can use Dijkstra's algorithm starting from the source node k. We will maintain a distance array to keep track of the minimum time it takes to reach each node. The algorithm involves relaxing edges and updating the distance array until all nodes are visited or we can't reach some nodes.

Initialize all distances to infinity except the source node k to 0.
Initialize a priority queue to store nodes based on their distances.
While the priority queue is not empty:
- Pop the node with the minimum distance.
- Relax all outgoing edges from this node, updating the distance array if a shorter path is found.
After processing all nodes, return the maximum value in the distance array if all nodes are reachable. Otherwise, return -1.

Time complexity: The time complexity of Dijkstra's algorithm is O((V + E) log V), where V is the number of nodes and E is the number of edges.

Space complexity: The space complexity is O(V) for storing the distance array and the priority queue.

Solution Code

class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        int[] dist = new int[n + 1];
        Arrays.fill(dist, Integer.MAX_VALUE);
        dist[k] = 0;
        
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[1] - b[1]);
        pq.offer(new int[]{k, 0});
        
        Map<Integer, List<int[]>> graph = new HashMap<>();
        for (int[] time : times) {
            graph.computeIfAbsent(time[0], x -> new ArrayList<>()).add(new int[]{time[1], time[2]});
        }
        
        while (!pq.isEmpty()) {
            int[] cur = pq.poll();
            int node = cur[0];
            int time = cur[1];
            
            if (!graph.containsKey(node)) continue;
            
            for (int[] next : graph.get(node)) {
                int neighbor = next[0];
                int newTime = time + next[1];
                
                if (newTime < dist[neighbor]) {
                    dist[neighbor] = newTime;
                    pq.offer(new int[]{neighbor, newTime});
                }
            }
        }
        
        int maxTime = Arrays.stream(dist).max().getAsInt();
        return maxTime == Integer.MAX_VALUE ? -1 : maxTime;
    }
}

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