Add solution #1514

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,340 LeetCode solutions in JavaScript
+# 1,341 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1157,6 +1157,7 @@
 1510|[Stone Game IV](./solutions/1510-stone-game-iv.js)|Hard|
 1512|[Number of Good Pairs](./solutions/1512-number-of-good-pairs.js)|Easy|
 1513|[Number of Substrings With Only 1s](./solutions/1513-number-of-substrings-with-only-1s.js)|Medium|
+1514|[Path with Maximum Probability](./solutions/1514-path-with-maximum-probability.js)|Medium|
 1519|[Number of Nodes in the Sub-Tree With the Same Label](./solutions/1519-number-of-nodes-in-the-sub-tree-with-the-same-label.js)|Medium|
 1524|[Number of Sub-arrays With Odd Sum](./solutions/1524-number-of-sub-arrays-with-odd-sum.js)|Medium|
 1528|[Shuffle String](./solutions/1528-shuffle-string.js)|Easy|

solutions/1514-path-with-maximum-probability.js

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+/**
+ * 1514. Path with Maximum Probability
+ * https://leetcode.com/problems/path-with-maximum-probability/
+ * Difficulty: Medium
+ *
+ * You are given an undirected weighted graph of n nodes (0-indexed), represented by an edge list
+ * where edges[i] = [a, b] is an undirected edge connecting the nodes a and b with a probability
+ * of success of traversing that edge succProb[i].
+ *
+ * Given two nodes start and end, find the path with the maximum probability of success to go from
+ * start to end and return its success probability.
+ *
+ * If there is no path from start to end, return 0. Your answer will be accepted if it differs from
+ * the correct answer by at most 1e-5.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {number[]} succProb
+ * @param {number} startNode
+ * @param {number} endNode
+ * @return {number}
+ */
+var maxProbability = function(n, edges, succProb, startNode, endNode) {
+  const maxProbs = new Array(n).fill(0);
+  maxProbs[startNode] = 1;
+
+  const graph = Array.from({ length: n }, () => []);
+  edges.forEach(([a, b], i) => {
+    graph[a].push([b, succProb[i]]);
+    graph[b].push([a, succProb[i]]);
+  });
+
+  const queue = [startNode];
+  while (queue.length) {
+    const current = queue.shift();
+    for (const [next, prob] of graph[current]) {
+      const newProb = maxProbs[current] * prob;
+      if (newProb > maxProbs[next]) {
+        maxProbs[next] = newProb;
+        queue.push(next);
+      }
+    }
+  }
+
+  return maxProbs[endNode];
+};