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            "questionFrontendId": "1514",
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            "title": "Path with Maximum Probability",
            "titleSlug": "path-with-maximum-probability",
            "content": "<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>\r\n\r\n<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>\r\n\r\n<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>\r\n\r\n<p>&nbsp;</p>\r\n<p><strong>Example 1:</strong></p>\r\n\r\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/20/1558_ex1.png\" style=\"width: 187px; height: 186px;\" /></strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2\r\n<strong>Output:</strong> 0.25000\r\n<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.\r\n</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n\r\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/20/1558_ex2.png\" style=\"width: 189px; height: 186px;\" /></strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2\r\n<strong>Output:</strong> 0.30000\r\n</pre>\r\n\r\n<p><strong>Example 3:</strong></p>\r\n\r\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/20/1558_ex3.png\" style=\"width: 215px; height: 191px;\" /></strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2\r\n<strong>Output:</strong> 0.00000\r\n<strong>Explanation:</strong>&nbsp;There is no path between 0 and 2.\r\n</pre>\r\n\r\n<p>&nbsp;</p>\r\n<p><strong>Constraints:</strong></p>\r\n\r\n<ul>\r\n\t<li><code>2 &lt;= n &lt;= 10^4</code></li>\r\n\t<li><code>0 &lt;= start, end &lt; n</code></li>\r\n\t<li><code>start != end</code></li>\r\n\t<li><code>0 &lt;= a, b &lt; n</code></li>\r\n\t<li><code>a != b</code></li>\r\n\t<li><code>0 &lt;= succProb.length == edges.length &lt;= 2*10^4</code></li>\r\n\t<li><code>0 &lt;= succProb[i] &lt;= 1</code></li>\r\n\t<li>There is at most one edge between every two nodes.</li>\r\n</ul>",
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                    "code": "class Solution {\npublic:\n    double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start, int end) {\n        \n    }\n};",
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                    "code": "class Solution {\n    public double maxProbability(int n, int[][] edges, double[] succProb, int start, int end) {\n        \n    }\n}",
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                    "code": "class Solution(object):\n    def maxProbability(self, n, edges, succProb, start, end):\n        \"\"\"\n        :type n: int\n        :type edges: List[List[int]]\n        :type succProb: List[float]\n        :type start: int\n        :type end: int\n        :rtype: float\n        \"\"\"\n        ",
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                    "code": "class Solution:\n    def maxProbability(self, n: int, edges: List[List[int]], succProb: List[float], start: int, end: int) -> float:\n        ",
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                    "lang": "C",
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                    "code": "\n\ndouble maxProbability(int n, int** edges, int edgesSize, int* edgesColSize, double* succProb, int succProbSize, int start, int end){\n\n}",
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                    "code": "public class Solution {\n    public double MaxProbability(int n, int[][] edges, double[] succProb, int start, int end) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number} n\n * @param {number[][]} edges\n * @param {number[]} succProb\n * @param {number} start\n * @param {number} end\n * @return {number}\n */\nvar maxProbability = function(n, edges, succProb, start, end) {\n    \n};",
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                    "code": "# @param {Integer} n\n# @param {Integer[][]} edges\n# @param {Float[]} succ_prob\n# @param {Integer} start\n# @param {Integer} end\n# @return {Float}\ndef max_probability(n, edges, succ_prob, start, end)\n    \nend",
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                    "lang": "Swift",
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                    "code": "class Solution {\n    func maxProbability(_ n: Int, _ edges: [[Int]], _ succProb: [Double], _ start: Int, _ end: Int) -> Double {\n        \n    }\n}",
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                    "code": "func maxProbability(n int, edges [][]int, succProb []float64, start int, end int) float64 {\n    \n}",
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                    "code": "object Solution {\n    def maxProbability(n: Int, edges: Array[Array[Int]], succProb: Array[Double], start: Int, end: Int): Double = {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun maxProbability(n: Int, edges: Array<IntArray>, succProb: DoubleArray, start: Int, end: Int): Double {\n        \n    }\n}",
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                    "lang": "Rust",
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                    "code": "impl Solution {\n    pub fn max_probability(n: i32, edges: Vec<Vec<i32>>, succ_prob: Vec<f64>, start: i32, end: i32) -> f64 {\n        \n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $n\n     * @param Integer[][] $edges\n     * @param Float[] $succProb\n     * @param Integer $start\n     * @param Integer $end\n     * @return Float\n     */\n    function maxProbability($n, $edges, $succProb, $start, $end) {\n        \n    }\n}",
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                    "lang": "TypeScript",
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                    "code": "function maxProbability(n: number, edges: number[][], succProb: number[], start: number, end: number): number {\n\n};",
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            "hints": [
                "Multiplying probabilities will result in precision errors.",
                "Take log probabilities to sum up numbers instead of multiplying them.",
                "Use Dijkstra's algorithm to find the minimum path between the two nodes after negating all costs."
            ],
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