Add solution #851

Author: JoshCrozier

README.md

@@ -658,6 +658,7 @@
 848|[Shifting Letters](./0848-shifting-letters.js)|Medium|
 849|[Maximize Distance to Closest Person](./0849-maximize-distance-to-closest-person.js)|Medium|
 850|[Rectangle Area II](./0850-rectangle-area-ii.js)|Hard|
+851|[Loud and Rich](./0851-loud-and-rich.js)|Medium|
 867|[Transpose Matrix](./0867-transpose-matrix.js)|Easy|
 868|[Binary Gap](./0868-binary-gap.js)|Easy|
 872|[Leaf-Similar Trees](./0872-leaf-similar-trees.js)|Easy|

solutions/0851-loud-and-rich.js

@@ -0,0 +1,53 @@
+/**
+ * 851. Loud and Rich
+ * https://leetcode.com/problems/loud-and-rich/
+ * Difficulty: Medium
+ *
+ * There is a group of n people labeled from 0 to n - 1 where each person has a different amount
+ * of money and a different level of quietness.
+ *
+ * You are given an array richer where richer[i] = [ai, bi] indicates that ai has more money than
+ * bi and an integer array quiet where quiet[i] is the quietness of the ith person. All the given
+ * data in richer are logically correct (i.e., the data will not lead you to a situation where x
+ * is richer than y and y is richer than x at the same time).
+ *
+ * Return an integer array answer where answer[x] = y if y is the least quiet person (that is,
+ * the person y with the smallest value of quiet[y]) among all people who definitely have equal
+ * to or more money than the person x.
+ */
+
+/**
+ * @param {number[][]} richer
+ * @param {number[]} quiet
+ * @return {number[]}
+ */
+var loudAndRich = function(richer, quiet) {
+  const graph = new Array(quiet.length).fill().map(() => []);
+  const result = new Array(quiet.length).fill(-1);
+
+  for (const [a, b] of richer) {
+    graph[b].push(a);
+  }
+
+  function dfs(person) {
+    if (result[person] !== -1) return result[person];
+
+    let quietestPerson = person;
+
+    for (const richerPerson of graph[person]) {
+      const candidate = dfs(richerPerson);
+      if (quiet[candidate] < quiet[quietestPerson]) {
+        quietestPerson = candidate;
+      }
+    }
+
+    result[person] = quietestPerson;
+    return quietestPerson;
+  }
+
+  for (let i = 0; i < quiet.length; i++) {
+    dfs(i);
+  }
+
+  return result;
+};