1514. Path with Maximum Probability - Explanation

Description

You are given an undirected weighted graph of n nodes (0-indexed), represented by an edge list where edges[i] = [a, b] is an undirected edge connecting the nodes a and b with a probability of success of traversing that edge succProb[i].

Given two nodes start and end, find the path with the maximum probability of success to go from start to end and return its success probability.

If there is no path from start to end, return 0. Your answer will be accepted if it differs from the correct answer by at most 1e-5.

Example 1:

Input: n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2

Output: 0.25000

Explanation: There are two paths from start to end, one having a probability of success = 0.2 and the other has 0.5 * 0.5 = 0.25.

Example 2:

Input: n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2

Output: 0.30000

Explanation: There is an edge which connects the nodes 0 and 2 with probability = 0.3.

Example 3:

Input: n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2

Output: 0.00000

Explanation: There is no path between 0 and 2.

Constraints:

2 <= n <= 10,000
0 <= start, end < n
start != end
0 <= a, b < n
a != b
0 <= succProb.length == edges.length <= 20,000
0 <= succProb[i] <= 1
There is at most one edge between every two nodes.

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1. Dijkstra's Algorithm - I

class Solution:
    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start_node: int, end_node: int) -> float:
        adj = collections.defaultdict(list)
        for i in range(len(edges)):
            src, dst = edges[i]
            adj[src].append((dst, succProb[i]))
            adj[dst].append((src, succProb[i]))

        pq = [(-1, start_node)]
        visit = set()

        while pq:
            prob, cur = heapq.heappop(pq)
            visit.add(cur)

            if cur == end_node:
                return -prob

            for nei, edgeProb in adj[cur]:
                if nei not in visit:
                    heapq.heappush(pq, (prob * edgeProb, nei))

        return 0.0

Time & Space Complexity

Time complexity: $O((V + E) \log V)$
Space complexity: $O(V + E)$

Where $V$ is the number nodes and $E$ is the number of edges.

2. Dijkstra's Algorithm - II

class Solution:
    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start_node: int, end_node: int) -> float:
        adj = [[] for _ in range(n)]
        for i in range(len(edges)):
            src, dst = edges[i]
            adj[src].append((dst, succProb[i]))
            adj[dst].append((src, succProb[i]))

        maxProb = [0] * n
        maxProb[start_node] = 1.0
        pq = [(-1.0, start_node)]

        while pq:
            curr_prob, node = heapq.heappop(pq)
            curr_prob *= -1

            if node == end_node:
                return curr_prob
            if curr_prob > maxProb[node]:
                continue

            for nei, edge_prob in adj[node]:
                new_prob = curr_prob * edge_prob
                if new_prob > maxProb[nei]:
                    maxProb[nei] = new_prob
                    heapq.heappush(pq, (-new_prob, nei))

        return 0.0

Time & Space Complexity

Time complexity: $O((V + E) \log V)$
Space complexity: $O(V + E)$

Where $V$ is the number nodes and $E$ is the number of edges.

3. Bellman Ford Algorithm

class Solution:
    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start_node: int, end_node: int) -> float:
        maxProb = [0.0] * n
        maxProb[start_node] = 1.0

        for i in range(n):
            updated = False
            for j in range(len(edges)):
                src, dst = edges[j]
                if maxProb[src] * succProb[j] > maxProb[dst]:
                    maxProb[dst] = maxProb[src] * succProb[j]
                    updated = True

                if maxProb[dst] * succProb[j] > maxProb[src]:
                    maxProb[src] = maxProb[dst] * succProb[j]
                    updated = True

            if not updated:
                break

        return maxProb[end_node]

Time & Space Complexity

Time complexity: $O(V * E)$
Space complexity: $O(V)$

Where $V$ is the number nodes and $E$ is the number of edges.

4. Shortest Path Faster Algorithm

class Solution:
    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start_node: int, end_node: int) -> float:
        adj = [[] for _ in range(n)]
        for i in range(len(edges)):
            src, dst = edges[i]
            adj[src].append((dst, succProb[i]))
            adj[dst].append((src, succProb[i]))

        maxProb = [0.0] * n
        maxProb[start_node] = 1.0
        q = deque([start_node])

        while q:
            node = q.popleft()

            for nei, edge_prob in adj[node]:
                new_prob = maxProb[node] * edge_prob
                if new_prob > maxProb[nei]:
                    maxProb[nei] = new_prob
                    q.append(nei)

        return maxProb[end_node]

Time & Space Complexity

Time complexity: $O(V * E)$
Space complexity: $O(V + E)$

Where $V$ is the number nodes and $E$ is the number of edges.