Add solution #1519

Author: JoshCrozier

README.md

@@ -285,6 +285,7 @@
 1502|[Can Make Arithmetic Progression From Sequence](./1502-can-make-arithmetic-progression-from-sequence.js)|Easy|
 1507|[Reformat Date](./1507-reformat-date.js)|Easy|
 1512|[Number of Good Pairs](./1512-number-of-good-pairs.js)|Easy|
+1519|[Number of Nodes in the Sub-Tree With the Same Label](./1519-number-of-nodes-in-the-sub-tree-with-the-same-label.js)|Medium|
 1528|[Shuffle String](./1528-shuffle-string.js)|Easy|
 1550|[Three Consecutive Odds](./1550-three-consecutive-odds.js)|Easy|
 1551|[Minimum Operations to Make Array Equal](./1551-minimum-operations-to-make-array-equal.js)|Medium|

solutions/1519-number-of-nodes-in-the-sub-tree-with-the-same-label.js

@@ -0,0 +1,52 @@
+/**
+ * 1519. Number of Nodes in the Sub-Tree With the Same Label
+ * https://leetcode.com/problems/number-of-nodes-in-the-sub-tree-with-the-same-label/
+ * Difficulty: Medium
+ *
+ * You are given a tree (i.e. a connected, undirected graph that has no cycles)
+ * consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. The
+ * root of the tree is the node 0, and each node of the tree has a label which
+ * is a lower-case character given in the string labels (i.e. The node with the
+ * number i has the label labels[i]).
+ *
+ * The edges array is given on the form edges[i] = [ai, bi], which means there
+ * is an edge between nodes ai and bi in the tree.
+ *
+ * Return an array of size n where ans[i] is the number of nodes in the subtree
+ * of the ith node which have the same label as node i.
+ *
+ * A subtree of a tree T is the tree consisting of a node in T and all of its
+ * descendant nodes.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {string} labels
+ * @return {number[]}
+ */
+var countSubTrees = function(n, edges, labels) {
+  const lookup = Array.from(Array(n), () => []);
+  const result = new Array(n).fill(0);
+
+  edges.forEach(([x, y]) => {
+    lookup[x].push(y);
+    lookup[y].push(x);
+  });
+
+  function bfs(index, previous, chars = new Array(26).fill(0)) {
+    const key = labels.charCodeAt(index) - 97;
+    const count = chars[key];
+    chars[key]++;
+    lookup[index].forEach(i => {
+      if (i !== previous) {
+        bfs(i, index, chars);
+      }
+    });
+    result[index] = chars[key] - count;
+  }
+
+  bfs(0, -1);
+
+  return result;
+};