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Author: jimit105

3645-maximize-the-number-of-target-nodes-after-connecting-trees-ii/README.md

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+<p>There exist two <strong>undirected </strong>trees with <code>n</code> and <code>m</code> nodes, labeled from <code>[0, n - 1]</code> and <code>[0, m - 1]</code>, respectively.</p>
+
+<p>You are given two 2D integer arrays <code>edges1</code> and <code>edges2</code> of lengths <code>n - 1</code> and <code>m - 1</code>, respectively, where <code>edges1[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the first tree and <code>edges2[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the second tree.</p>
+
+<p>Node <code>u</code> is <strong>target</strong> to node <code>v</code> if the number of edges on the path from <code>u</code> to <code>v</code> is even.&nbsp;<strong>Note</strong> that a node is <em>always</em> <strong>target</strong> to itself.</p>
+
+<p>Return an array of <code>n</code> integers <code>answer</code>, where <code>answer[i]</code> is the <strong>maximum</strong> possible number of nodes that are <strong>target</strong> to node <code>i</code> of the first tree if you had to connect one node from the first tree to another node in the second tree.</p>
+
+<p><strong>Note</strong> that queries are independent from each other. That is, for every query you will remove the added edge before proceeding to the next query.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">edges1 = [[0,1],[0,2],[2,3],[2,4]], edges2 = [[0,1],[0,2],[0,3],[2,7],[1,4],[4,5],[4,6]]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[8,7,7,8,8]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For <code>i = 0</code>, connect node 0 from the first tree to node 0 from the second tree.</li>
+	<li>For <code>i = 1</code>, connect node 1 from the first tree to node 4 from the second tree.</li>
+	<li>For <code>i = 2</code>, connect node 2 from the first tree to node 7 from the second tree.</li>
+	<li>For <code>i = 3</code>, connect node 3 from the first tree to node 0 from the second tree.</li>
+	<li>For <code>i = 4</code>, connect node 4 from the first tree to node 4 from the second tree.</li>
+</ul>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/09/24/3982-1.png" style="width: 600px; height: 169px;" /></div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">edges1 = [[0,1],[0,2],[0,3],[0,4]], edges2 = [[0,1],[1,2],[2,3]]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[3,6,6,6,6]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>For every <code>i</code>, connect node <code>i</code> of the first tree with any node of the second tree.</p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/09/24/3928-2.png" style="height: 281px; width: 500px;" /></div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n, m &lt;= 10<sup>5</sup></code></li>
+	<li><code>edges1.length == n - 1</code></li>
+	<li><code>edges2.length == m - 1</code></li>
+	<li><code>edges1[i].length == edges2[i].length == 2</code></li>
+	<li><code>edges1[i] = [a<sub>i</sub>, b<sub>i</sub>]</code></li>
+	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
+	<li><code>edges2[i] = [u<sub>i</sub>, v<sub>i</sub>]</code></li>
+	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; m</code></li>
+	<li>The input is generated such that <code>edges1</code> and <code>edges2</code> represent valid trees.</li>
+</ul>

3645-maximize-the-number-of-target-nodes-after-connecting-trees-ii/solution.py

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+class Solution:
+    def maxTargetNodes(
+        self, edges1: List[List[int]], edges2: List[List[int]]
+    ) -> List[int]:
+        def dfs(node, parent, depth, children, color):
+            res = 1 - depth % 2
+            color[node] = depth % 2
+            for child in children[node]:
+                if child == parent:
+                    continue
+                res += dfs(child, node, depth + 1, children, color)
+            return res
+
+        def build(edges, color):
+            n = len(edges) + 1
+            children = [[] for _ in range(n)]
+            for u, v in edges:
+                children[u].append(v)
+                children[v].append(u)
+            res = dfs(0, -1, 0, children, color)
+            return [res, n - res]
+
+        n = len(edges1) + 1
+        m = len(edges2) + 1
+        color1 = [0] * n
+        color2 = [0] * m
+        count1 = build(edges1, color1)
+        count2 = build(edges2, color2)
+        res = [0] * n
+        for i in range(n):
+            res[i] = count1[color1[i]] + max(count2[0], count2[1])
+        return res