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

<p>Given two nodes&nbsp;<code>start</code>&nbsp;and&nbsp;<code>end</code>, find the path with the maximum probability of success to go from&nbsp;<code>start</code>&nbsp;to&nbsp;<code>end</code>&nbsp;and return its success probability.</p>

<p>If there is no path from&nbsp;<code>start</code>&nbsp;to&nbsp;<code>end</code>, <strong>return&nbsp;0</strong>. Your answer will be accepted if it differs from the correct answer by at most <strong>1e-5</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/20/1558_ex1.png" style="width: 187px; height: 186px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2
<strong>Output:</strong> 0.25000
<strong>Explanation:</strong>&nbsp;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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/20/1558_ex2.png" style="width: 189px; height: 186px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2
<strong>Output:</strong> 0.30000
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/20/1558_ex3.png" style="width: 215px; height: 191px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2
<strong>Output:</strong> 0.00000
<strong>Explanation:</strong>&nbsp;There is no path between 0 and 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10^4</code></li>
	<li><code>0 &lt;= start, end &lt; n</code></li>
	<li><code>start != end</code></li>
	<li><code>0 &lt;= a, b &lt; n</code></li>
	<li><code>a != b</code></li>
	<li><code>0 &lt;= succProb.length == edges.length &lt;= 2*10^4</code></li>
	<li><code>0 &lt;= succProb[i] &lt;= 1</code></li>
	<li>There is at most one edge between every two nodes.</li>
</ul>

#include <iostream>
#include <vector>
#include <queue>
#include <limits>
#include <utility> // For std::pair
using namespace std;

typedef pair<int, double> ii;
typedef long long ll;
#define F first
#define S second

vector<ii> g[100100]; // Graph adjacency list
double prob[100100];  // Probability array

class prioritize {
public:
    bool operator()(ii &p1, ii &p2) {
        return p1.S < p2.S; // Max-heap based on probability (greater comes first)
    }
};

class Solution {
public:
    double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start_node, int end_node) {
        // Clear the graph and probability array for multiple test cases
        for(int i = 0; i < n; i++) {
            g[i].clear();
            prob[i] = 0.0;
        }
        
        // Construct the graph
        for(int i = 0; i < edges.size(); i++) {
            g[edges[i][0]].emplace_back(edges[i][1], succProb[i]);
            g[edges[i][1]].emplace_back(edges[i][0], succProb[i]);
        }
        
        prob[start_node] = 1.0; // Start node has probability 1

        // Priority queue to process nodes by maximum probability
        priority_queue<ii, vector<ii>, prioritize> pq;
        pq.push({start_node, 1.0});
        
        while(!pq.empty()) {
            ii current = pq.top();
            pq.pop();
            int u = current.F;
            double curr_prob = current.S;
            
            if (u == end_node) {
                return curr_prob; // Return the max probability when the end node is reached
            }

            // Traverse adjacent nodes
            for(auto &v : g[u]) {
                int neigh = v.F;
                double edge_prob = v.S;
                
                // If a better probability path is found, update and push to the queue
                if (prob[neigh] < curr_prob * edge_prob) {
                    prob[neigh] = curr_prob * edge_prob;
                    pq.push({neigh, prob[neigh]});
                }
            }
        }
        
        return 0.0; // If no path is found, return 0.0
    }
};


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