Find Minimum Diameter After Merging Two Trees

Tested Python solution for LeetCode 3203 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 3203, Hard. Topics: Tree, Depth-First Search, Breadth-First Search, Graph Theory. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 18 parametrized pytest cases, and a playground notebook:

lcpy gen -n 3203   # by problem number
lcpy gen -s find_minimum_diameter_after_merging_two_trees   # by problem name

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Problem

There exist two undirected trees with n and m nodes, numbered from 0 to n - 1 and from 0 to 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.

You must connect one node from the first tree with another node from the second tree with an edge.

Return the minimum possible diameter of the resulting tree.

The diameter of a tree is the length of the longest path between any two nodes in the tree.

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Examples

Input: edges1 = [[0,1],[0,2],[0,3]], edges2 = [[0,1]]
Output: 3

Explanation: We can obtain a tree of diameter 3 by connecting node 0 from the first tree with any node from the second tree.

Input: 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]]
Output: 5

Explanation: We can obtain a tree of diameter 5 by connecting node 0 from the first tree with node 0 from the second tree.

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Constraints

• 1 <= n, m <= 10^5
• edges1.length == n - 1
• edges2.length == m - 1
• edges1[i].length == edges2[i].length == 2
• edges1[i] = [ai, bi]
• 0 <= ai, bi < n
• edges2[i] = [ui, vi]
• 0 <= ui, vi < m
• The input is generated such that edges1 and edges2 represent valid trees.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from collections import deque


class Solution:
    # Time: O(n + m)
    # Space: O(n + m)
    def minimum_diameter_after_merge(self, edges1: list[list[int]], edges2: list[list[int]]) -> int:
        d1 = self._diameter(len(edges1) + 1, edges1)
        d2 = self._diameter(len(edges2) + 1, edges2)
        return max(d1, d2, (d1 + 1) // 2 + (d2 + 1) // 2 + 1)

    def _diameter(self, n: int, edges: list[list[int]]) -> int:
        adj: list[list[int]] = [[] for _ in range(n)]
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)

        def farthest(src: int) -> tuple[int, int]:
            dist = [-1] * n
            dist[src] = 0
            queue = deque([src])
            last = src
            while queue:
                node = queue.popleft()
                last = node
                for nxt in adj[node]:
                    if dist[nxt] == -1:
                        dist[nxt] = dist[node] + 1
                        queue.append(nxt)
            return last, dist[last]

        endpoint, _ = farthest(0)
        _, diameter = farthest(endpoint)
        return diameter

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Complexity

Time	Space
O(n + m)	O(n + m)

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Tags

NeetCode All.