743-network-delay-time

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Description

You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.

We will send a signal from a given node k. Return the time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

 

Example 1:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2

Example 2:

Input: times = [[1,2,1]], n = 2, k = 1
Output: 1

Example 3:

Input: times = [[1,2,1]], n = 2, k = 2
Output: -1

 

Constraints:

• 1 <= k <= n <= 100
• 1 <= times.length <= 6000
• times[i].length == 3
• 1 <= ui, vi <= n
• ui != vi
• 0 <= wi <= 100
• All the pairs (ui, vi) are unique. (i.e., no multiple edges.)

Solution(Python)

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        return self.dijikstra(times, n, k)

    # Time Complexity: O((N-1)!+E log E)
    # Space COmplexity: O(N+E)
    def dfs(self, times: List[List[int]], n: int, k: int) -> int:
        adj = defaultdict(list)
        for x, y, z in times:
            adj[x].append((y, z))

        times.sort(key=lambda time: time[2])
        signalReceivedAt = [sys.maxsize] * (n)

        def recur(node, currTime):
            if currTime >= signalReceivedAt[node - 1]:
                return
            signalReceivedAt[node - 1] = currTime
            for y, w in adj[node]:
                recur(y, currTime + w)

        recur(k, 0)

        max_ = max(signalReceivedAt)
        return max_ if max_ < sys.maxsize else -1

    # Time Complexity: O(N*E)
    # Space COmplexity: O(N+E)
    def bfs(self, times: List[List[int]], n: int, k: int) -> int:
        adj = defaultdict(list)
        for x, y, z in times:
            adj[x].append((y, z))

        signalReceivedAt = [sys.maxsize] * (n)
        q = deque([k])
        signalReceivedAt[k - 1] = 0
        while q:
            node = q.popleft()
            for y, w in adj[node]:
                arrivalTime = signalReceivedAt[node - 1] + w
                if arrivalTime < signalReceivedAt[y - 1]:
                    signalReceivedAt[y - 1] = arrivalTime
                    q.append(y)

        max_ = max(signalReceivedAt)
        return max_ if max_ < sys.maxsize else -1

    # Time Complexity: O(N+ElogE)
    # Space COmplexity: O(N+E)
    def dijikstra(self, times: List[List[int]], n: int, k: int) -> int:
        adj = defaultdict(list)
        for x, y, z in times:
            adj[x].append((y, z))

        visited = set()
        pq = [(0, k)]
        heapq.heapify(pq)
        maxcost = 0
        while pq:
            cost, node = heapq.heappop(pq)
            if node in visited:
                continue
            if cost > maxcost:
                maxcost = cost
            visited.add(node)

            for y, w in adj[node]:
                newcost = cost + w
                if y not in visited:
                    heapq.heappush(pq, (newcost, y))

        return maxcost if n == len(visited) else -1