**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. Constraints 1 ≤ n ≤ 100,000 where n is the length of edges and success 0 ≤ success[i] ≤ 1 Example 1 Input edges = [ [0, 1], [1, 2], [2, 3] ] success = [0.2, 0.2, 0.5] start = 0 end = 3 Output 0.02 Explanation 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] = pq.top();
pq.pop();
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:
continue
visited.add(i)
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
```

## View More Similar Problems

## AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

View Solution →## Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

View Solution →## Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →## Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →## Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →## Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →