#3373

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Maximize the Number of Target Nodes After Connecting Trees II

Maximize the number of target nodes after connecting two trees by analyzing their structure and target relationships.

tree depth first search breadth first search

Problem Statement

You are given two undirected trees with n and m nodes. Each tree is defined by edges where the nodes are labeled from 0 to n-1 for the first tree and 0 to m-1 for the second. Your task is to connect nodes from the first tree to nodes in the second tree in a way that maximizes the number of 'target' nodes. A node u is a target to node v if the number of edges between them is even, and a node is always a target to itself.

You are provided two 2D arrays, edges1 and edges2, representing the edges of each tree. You need to compute the number of target nodes that can be reached by connecting nodes from the first tree to nodes in the second tree. The challenge requires efficient binary-tree traversal and managing the state of distances to identify the maximum target connections.

Examples

Example 1

Input: edges1 = [[0,1],[0,2],[2,3],[2,4]], edges2 = [[0,1],[0,2],[0,3],[2,7],[1,4],[4,5],[4,6]]

Output: [8,7,7,8,8]

Example 2

Input: edges1 = [[0,1],[0,2],[0,3],[0,4]], edges2 = [[0,1],[1,2],[2,3]]

Output: [3,6,6,6,6]

For every i , connect node i of the first tree with any node of the second tree.

Constraints

2 <= n, m <= 105
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.

Solution Approach

Tree Traversal and Distance Calculation

To solve the problem, first compute the distance of each node in both trees using binary-tree traversal techniques such as Depth-First Search (DFS). Track even and odd distances from each node in both trees, which will help to determine the number of target nodes reachable from any given node.

State Tracking for Even Distances

Maintain an array where each element tracks the number of nodes at an even distance from the current node. Use this array to identify which nodes can be connected to form target pairs. After calculating the even distance nodes in both trees, the solution will be based on matching nodes that form even distances when connected.

Efficient Node Matching

To maximize the number of target nodes, efficiently match the nodes of the two trees based on their even-distance properties. The optimal solution requires iterating over all possible node connections and selecting those that maximize the number of target nodes while keeping the overall complexity manageable.

Complexity Analysis

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

The time complexity of this solution is O(n + m), where n is the number of nodes in the first tree and m is the number of nodes in the second tree. The space complexity is also O(n + m) due to the need for storing node distances and state information during the traversal and matching process.

What Interviewers Usually Probe

Check the candidate's ability to traverse trees and compute distances efficiently.
Evaluate their understanding of target node relationships based on even and odd distances.
Assess their ability to optimize the solution using efficient data structures and algorithms.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly track the even and odd distances during traversal can lead to incorrect target node calculations.
Not optimizing the solution for large inputs (n and m up to 10^5) may result in a time-out.
Overlooking edge cases where trees are highly imbalanced or when the connection strategy doesn't lead to maximizing the target nodes.

Follow-up variants

Change the problem by considering directed trees instead of undirected trees.
Consider other types of node relationships, such as distance-based instead of even-odd distance.
Modify the problem to calculate the number of nodes that are reachable based on a different distance metric (e.g., odd distance).

FAQ

How can I maximize the number of target nodes in this problem?

To maximize the target nodes, you need to efficiently calculate and match nodes with even distances from each other across the two trees.

What traversal techniques are required to solve the Maximize the Number of Target Nodes problem?

The solution involves using Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the trees and calculate distances between nodes.

What is the primary algorithmic pattern for this problem?

The primary algorithmic pattern is binary-tree traversal combined with state tracking of even and odd distances to maximize target nodes.

What are the space and time complexities of the solution?

The time complexity is O(n + m) and the space complexity is O(n + m), where n and m are the sizes of the two trees.

How can I handle large inputs efficiently in this problem?

Efficient traversal and state tracking with careful optimization are key to handling large inputs without exceeding time limits.

terminal

Solution

Solution 1: DFS

The number of target nodes for node $i$ can be divided into two parts:

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class Solution:
    def maxTargetNodes(
        self, edges1: List[List[int]], edges2: List[List[int]]
    ) -> List[int]:
        def build(edges: List[List[int]]) -> List[List[int]]:
            n = len(edges) + 1
            g = [[] for _ in range(n)]
            for a, b in edges:
                g[a].append(b)
                g[b].append(a)
            return g

        def dfs(
            g: List[List[int]], a: int, fa: int, c: List[int], d: int, cnt: List[int]
        ):
            c[a] = d
            cnt[d] += 1
            for b in g[a]:
                if b != fa:
                    dfs(g, b, a, c, d ^ 1, cnt)

        g1 = build(edges1)
        g2 = build(edges2)
        n, m = len(g1), len(g2)
        c1 = [0] * n
        c2 = [0] * m
        cnt1 = [0, 0]
        cnt2 = [0, 0]
        dfs(g2, 0, -1, c2, 0, cnt2)
        dfs(g1, 0, -1, c1, 0, cnt1)
        t = max(cnt2)
        return [t + cnt1[c1[i]] for i in range(n)]

Continue Practicing

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