Add solution #3373

Author: JoshCrozier

README.md

@@ -2044,6 +2044,7 @@
 3356|[Zero Array Transformation II](./solutions/3356-zero-array-transformation-ii.js)|Medium|
 3362|[Zero Array Transformation III](./solutions/3362-zero-array-transformation-iii.js)|Medium|
 3372|[Maximize the Number of Target Nodes After Connecting Trees I](./solutions/3372-maximize-the-number-of-target-nodes-after-connecting-trees-i.js)|Medium|
+3373|[Maximize the Number of Target Nodes After Connecting Trees II](./solutions/3373-maximize-the-number-of-target-nodes-after-connecting-trees-ii.js)|Hard|
 3375|[Minimum Operations to Make Array Values Equal to K](./solutions/3375-minimum-operations-to-make-array-values-equal-to-k.js)|Easy|
 3392|[Count Subarrays of Length Three With a Condition](./solutions/3392-count-subarrays-of-length-three-with-a-condition.js)|Easy|
 3394|[Check if Grid can be Cut into Sections](./solutions/3394-check-if-grid-can-be-cut-into-sections.js)|Medium|

solutions/3373-maximize-the-number-of-target-nodes-after-connecting-trees-ii.js

@@ -0,0 +1,79 @@
+/**
+ * 3373. Maximize the Number of Target Nodes After Connecting Trees II
+ * https://leetcode.com/problems/maximize-the-number-of-target-nodes-after-connecting-trees-ii/
+ * Difficulty: Hard
+ *
+ * There exist two undirected trees with n and m nodes, labeled from [0, n - 1] and [0, m - 1],
+ * respectively.
+ *
+ * You are given two 2D integer arrays edges1 and edges2 of lengths n - 1 and m - 1, respectively,
+ * where edges1[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the first
+ * tree and edges2[i] = [ui, vi] indicates that there is an edge between nodes ui and vi in the
+ * second tree.
+ *
+ * Node u is target to node v if the number of edges on the path from u to v is even. Note that
+ * a node is always target to itself.
+ *
+ * Return an array of n integers answer, where answer[i] is the maximum possible number of nodes
+ * that are target to node i of the first tree if you had to connect one node from the first tree
+ * to another node in the second tree.
+ *
+ * Note that queries are independent from each other. That is, for every query you will remove the
+ * added edge before proceeding to the next query.
+ */
+
+/**
+ * @param {number[][]} edges1
+ * @param {number[][]} edges2
+ * @return {number[]}
+ */
+var maxTargetNodes = function(edges1, edges2) {
+  const graph1 = buildGraph(edges1);
+  const graph2 = buildGraph(edges2);
+  const n = graph1.length;
+  const { color: color1 } = getBipartiteGroups(graph1);
+  const { maxGroup: maxGroup2 } = getBipartiteGroups(graph2);
+  const group0Count = color1.filter(c => c === 0).length;
+  const group1Count = n - group0Count;
+  const result = new Array(n);
+
+  for (let i = 0; i < n; i++) {
+    result[i] = (color1[i] === 0 ? group0Count : group1Count) + maxGroup2;
+  }
+
+  return result;
+
+  function buildGraph(edges) {
+    const n = edges.length + 1;
+    const graph = Array.from({ length: n }, () => []);
+    for (const [u, v] of edges) {
+      graph[u].push(v);
+      graph[v].push(u);
+    }
+    return graph;
+  }
+
+  function getBipartiteGroups(graph) {
+    const n = graph.length;
+    const color = new Array(n).fill(-1);
+    const groups = [0, 0];
+
+    for (let i = 0; i < n; i++) {
+      if (color[i] === -1) {
+        dfs(i, 0);
+      }
+    }
+
+    return { color, maxGroup: Math.max(groups[0], groups[1]) };
+
+    function dfs(node, c) {
+      color[node] = c;
+      groups[c]++;
+      for (const neighbor of graph[node]) {
+        if (color[neighbor] === -1) {
+          dfs(neighbor, 1 - c);
+        }
+      }
+    }
+  }
+};