Add solution #834

Author: JoshCrozier

README.md

@@ -641,6 +641,7 @@
 830|[Positions of Large Groups](./0830-positions-of-large-groups.js)|Easy|
 831|[Masking Personal Information](./0831-masking-personal-information.js)|Medium|
 833|[Find And Replace in String](./0833-find-and-replace-in-string.js)|Medium|
+834|[Sum of Distances in Tree](./0834-sum-of-distances-in-tree.js)|Hard|
 841|[Keys and Rooms](./0841-keys-and-rooms.js)|Medium|
 844|[Backspace String Compare](./0844-backspace-string-compare.js)|Easy|
 846|[Hand of Straights](./0846-hand-of-straights.js)|Medium|

solutions/0834-sum-of-distances-in-tree.js

@@ -0,0 +1,53 @@
+/**
+ * 834. Sum of Distances in Tree
+ * https://leetcode.com/problems/sum-of-distances-in-tree/
+ * Difficulty: Hard
+ *
+ * There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.
+ *
+ * You are given the integer n and the array edges where edges[i] = [ai, bi] indicates that there
+ * is an edge between nodes ai and bi in the tree.
+ *
+ * Return an array answer of length n where answer[i] is the sum of the distances between the ith
+ * node in the tree and all other nodes.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @return {number[]}
+ */
+var sumOfDistancesInTree = function(n, edges) {
+  const graph = Array.from({ length: n }, () => []);
+  const count = new Array(n).fill(1);
+  const result = new Array(n).fill(0);
+
+  for (const [a, b] of edges) {
+    graph[a].push(b);
+    graph[b].push(a);
+  }
+
+  function dfs1(node, parent) {
+    for (const child of graph[node]) {
+      if (child !== parent) {
+        dfs1(child, node);
+        count[node] += count[child];
+        result[node] += result[child] + count[child];
+      }
+    }
+  }
+
+  function dfs2(node, parent) {
+    for (const child of graph[node]) {
+      if (child !== parent) {
+        result[child] = result[node] - count[child] + (n - count[child]);
+        dfs2(child, node);
+      }
+    }
+  }
+
+  dfs1(0, -1);
+  dfs2(0, -1);
+
+  return result;
+};