Published: Oct 21, 2022
It’s apparent, Dijkstra algorithm works to find the answer. After creating a graph with edge weight, traverse the graph using heap. When the node is the end node, the result is there.
You are given an undirected weighted graph of
nnodes (0-indexed), represented by an edge list where
edges[i] = [a, b]is an undirected edge connecting the nodes
bwith a probability of success of traversing that edge
Given two nodes
end, find the path with the maximum probability of success to go from
endand return its success probability.
If there is no path from
end, return 0. Your answer will be accepted if it differs from the correct answer by at most 1e-5.
2 <= n <= 10^4
0 <= start, end < n
start != end
0 <= a, b < n
a != b
0 <= succProb.length == edges.length <= 2*10^4
0 <= succProb[i] <= 1
- There is at most one edge between every two nodes.
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
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.
The solution here uses Python’s collections.defaultdict for isolated node (no edges) to return empty array. The adjacency list has the neighbor node with edge weight. Once the graph is created, the next step is to traverse the graph. To make it max heap, the probability is a negative value. Except probability calculation, traversal is nothing special. When popped out node is the end node, return the probability making it positive.
class PathWithMaximumProbability: def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start: int, end: int) -> float: graph = collections.defaultdict(list) def buildGraph(): for i in range(len(edges)): s, d, v = edges[i], edges[i], succProb[i] graph[s].append((d, v)) graph[d].append((s, v)) def search(): visited = set() queue = [(-1, start)] while queue: cur_prob, cur_n = heapq.heappop(queue) if cur_n == end: return (-1) * cur_prob visited.add(cur_n) for next_n, next_prob in graph[cur_n]: if next_n not in visited: heapq.heappush(queue, (next_prob * cur_prob, next_n)) return 0 buildGraph() return search()
O(e * log(v))– v: number of vertices, n: number of edges on each vertex
O(e + v)