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Author: NamVr

3439-find-minimum-diameter-after-merging-two-trees/README.md

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+<p>There exist two <strong>undirected </strong>trees with <code>n</code> and <code>m</code> nodes, numbered from <code>0</code> to <code>n - 1</code> and from <code>0</code> to <code>m - 1</code>, respectively. 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>You must connect one node from the first tree with another node from the second tree with an edge.</p>
+
+<p>Return the <strong>minimum </strong>possible <strong>diameter </strong>of the resulting tree.</p>
+
+<p>The <strong>diameter</strong> of a tree is the length of the <em>longest</em> path between any two nodes in the tree.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong><img alt="" src="https://assets.leetcode.com/uploads/2024/04/22/example11-transformed.png" /></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">edges1 = [[0,1],[0,2],[0,3]], edges2 = [[0,1]]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can obtain a tree of diameter 3 by connecting node 0 from the first tree with any node from the second tree.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/04/22/example211.png" />
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">edges1 = [[0,1],[0,2],[0,3],[2,4],[2,5],[3,6],[2,7]], edges2 = [[0,1],[0,2],[0,3],[2,4],[2,5],[3,6],[2,7]]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can obtain a tree of diameter 5 by connecting node 0 from the first tree with node 0 from the second tree.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &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>

3439-find-minimum-diameter-after-merging-two-trees/solution.py

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+class Solution:
+    def minimumDiameterAfterMerge(self, edges1: List[List[int]], edges2: List[List[int]]) -> int:
+        n = len(edges1) + 1
+        m = len(edges2) + 1
+
+        def build_adj(edges):
+            adj = defaultdict(list) # creates a dictionary of empty lists
+            for n1,n2 in edges:
+                adj[n1].append(n2)
+                adj[n2].append(n1)
+            return adj
+
+        adj1 = build_adj(edges1)
+        adj2 = build_adj(edges2)
+
+        # DFS : Get diameter
+        # return: (diam, max_left_path)
+        def get_diameter(cur, par, adj):
+            max_d = 0
+            max_child_paths = [0,0]
+
+            for nei in adj[cur]:
+                if nei == par:
+                    continue
+                
+                nei_d, nei_max_leaf_path = get_diameter(nei, cur, adj)
+                max_d = max(max_d, nei_d)
+
+                heappush(max_child_paths, nei_max_leaf_path)
+                heappop(max_child_paths) # min heap gets the minimum and pop it
+
+            max_d = max(max_d, sum(max_child_paths))
+            return [max_d, 1 + max(max_child_paths)]
+        
+        # diameters
+        d1, _ = get_diameter(0, -1, adj1)
+        d2, _ = get_diameter(0, -1, adj2)
+
+        return max(d1, d2, ceil(d1/2) + ceil(d2/2) + 1)