Path to Maximize Probability to Destination - Google Top Interview Questions

Problem Statement :

You are given a two-dimensional list of integers edges, a list of floats success, and integers start and end. edges represents an undirected, weighted graph. 

Each edges[i] contains [u, v] meaning u and v is connected, and the chance of successfully traversing the edge is success[i] chance.

Given that you start at node start and you want to arrive at node end, return the highest overall chance you can get by taking any path. If there's no path to end, return 0.


1 ≤ n ≤ 100,000 where n is the length of edges and success

0 ≤ success[i] ≤ 1

Example 1


edges = [

    [0, 1],

    [1, 2],

    [2, 3]


success = [0.2, 0.2, 0.5]

start = 0

end = 3




The probability of being able to go from 0 to 3 is 0.2 * 0.2 * 0.5.

Solution :


                        Solution in C++ :

double solve(vector<vector<int>>& edges, vector<double>& success, int start, int end) {
    unordered_map<int, vector<pair<int, double>>> adj;
    for (int i = 0; i < edges.size(); ++i) {
        int u = edges[i][0], v = edges[i][1];
        double w = success[i];
        adj[u].emplace_back(v, w);
        adj[v].emplace_back(u, w);
    unordered_map<int, double> dist;
    dist[start] = 1;
    priority_queue<pair<double, int>> pq;
    pq.emplace(1, start);
    while (!pq.empty()) {
        auto [cw, u] =;
        if (cw < dist[u]) continue;
        for (auto [v, w] : adj[u]) {
            if (cw * w > dist[v]) {
                dist[v] = cw * w;
                pq.emplace(dist[v], v);
    return dist[end];

                        Solution in Java :

import java.util.*;

class Solution {
    class State {
        int node;
        double prob;
        State(int _node, double _prob) {
            node = _node;
            prob = _prob;

    public double solve(int[][] edges, double[] success, int start, int end) {
        Map<Integer, List<State>> map = new HashMap<>();
        Map<Integer, Double> bestProb = new HashMap<>();
        for (int i = 0; i < edges.length; ++i) {
            // Add u, v to graphMap. Add v to u's list of neighbors & vice versa
            int[] edge = edges[i];

            map.putIfAbsent(edge[0], new ArrayList<>());
            map.putIfAbsent(edge[1], new ArrayList<>());

            map.get(edge[0]).add(new State(edge[1], success[i]));
            map.get(edge[1]).add(new State(edge[0], success[i]));

            // Set best prob. of each node to 0
            bestProb.putIfAbsent(edge[0], 0.0);
            bestProb.putIfAbsent(edge[1], 0.0);

        // Use PriorityQueue that sorts by max probability, add Start node with 100% probability
        Comparator<State> nodeSort = (a, b) -> (((Double) b.prob).compareTo((Double) a.prob));
        PriorityQueue<State> pq = new PriorityQueue<>(nodeSort);
        pq.add(new State(start, 1.0));

        // Begin Dijkstra's
        while (!pq.isEmpty()) {
            // Remove node from PQ
            State state = pq.poll();
            int parent = state.node;
            double prob = state.prob;

            // Reached goal!
            if (parent == end)
                return prob;

            // Explore neighbors
            for (State child : map.getOrDefault(parent, new ArrayList<>())) {
                // Get child's value, prob to visit it, & bestProb
                int childVal = child.node;
                double childProb = child.prob;
                double bestChildProb = bestProb.get(childVal);

                // Add to pq only if it can make a better prob
                if (bestChildProb < prob * childProb) {
                    pq.add(new State(childVal, prob * childProb));
                    bestProb.put(childVal, prob * childProb);

        return 0;

                        Solution in Python : 
class Solution:
    def solve(self, edges, success, start, end):
        adj = {}
        distances = {}
        for i, edge in enumerate(edges):
            a, b = edge
            adj[a] = adj.get(a, {})
            adj[b] = adj.get(b, {})
            adj[a][b] = success[i]
            adj[b][a] = success[i]
            distances[a] = 0
            distances[b] = 0

        distances[start] = 1
        heap = [(-1, start)]
        visited = set()

        while heap:
            prob, i = heapq.heappop(heap)
            prob = -prob
            if i in visited:

            if i == end:
                return prob

            for key, val in adj[i].items():
                if prob * val > distances[key]:
                    distances[key] = prob * val
                    heapq.heappush(heap, (-distances[key], key))

        return 0

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