LeetCode #3373 — HARD

Maximize the Number of Target Nodes After Connecting Trees II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the first tree and edges2[i] = [u_i, v_i] indicates that there is an edge between nodes u_i and v_i 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.

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]

Explanation:

For i = 0, connect node 0 from the first tree to node 0 from the second tree.
For i = 1, connect node 1 from the first tree to node 4 from the second tree.
For i = 2, connect node 2 from the first tree to node 7 from the second tree.
For i = 3, connect node 3 from the first tree to node 0 from the second tree.
For i = 4, connect node 4 from the first tree to node 4 from the second tree.

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]

Explanation:

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] = [a_i, b_i]
0 <= a_i, b_i < n
edges2[i] = [u_i, v_i]
0 <= u_i, v_i < m
The input is generated such that edges1 and edges2 represent valid trees.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[[0,1],[0,2],[2,3],[2,4]]
[[0,1],[0,2],[0,3],[2,7],[1,4],[4,5],[4,6]]

Example 2

[[0,1],[0,2],[0,3],[0,4]]
[[0,1],[1,2],[2,3]]

Core Insight

What unlocks the optimal approach

Compute an array <code>even</code> where <code>even[u]</code> is the number of nodes at an even distance from node <code>u</code>, for every <code>u</code> of the first tree.
Compute an array <code>odd</code> where <code>odd[u]</code> is the number of nodes at an odd distance from node <code>u</code>, for every <code>u</code> of the second tree.
<code>answer[i] = even[i]+ max(odd[1], odd[2], …, odd[m - 1])</code>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3373: Maximize the Number of Target Nodes After Connecting Trees II
class Solution {
    public int[] maxTargetNodes(int[][] edges1, int[][] edges2) {
        var g1 = build(edges1);
        var g2 = build(edges2);
        int n = g1.length, m = g2.length;
        int[] c1 = new int[n];
        int[] c2 = new int[m];
        int[] cnt1 = new int[2];
        int[] cnt2 = new int[2];
        dfs(g2, 0, -1, c2, 0, cnt2);
        dfs(g1, 0, -1, c1, 0, cnt1);
        int t = Math.max(cnt2[0], cnt2[1]);
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = t + cnt1[c1[i]];
        }
        return ans;
    }

    private List<Integer>[] build(int[][] edges) {
        int n = edges.length + 1;
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        return g;
    }

    private void dfs(List<Integer>[] g, int a, int fa, int[] c, int d, int[] cnt) {
        c[a] = d;
        cnt[d]++;
        for (int b : g[a]) {
            if (b != fa) {
                dfs(g, b, a, c, d ^ 1, cnt);
            }
        }
    }
}

// Accepted solution for LeetCode #3373: Maximize the Number of Target Nodes After Connecting Trees II
func maxTargetNodes(edges1 [][]int, edges2 [][]int) []int {
    g1 := build(edges1)
    g2 := build(edges2)
    n, m := len(g1), len(g2)
    c1 := make([]int, n)
    c2 := make([]int, m)
    cnt1 := make([]int, 2)
    cnt2 := make([]int, 2)

    dfs(g2, 0, -1, c2, 0, cnt2)
    dfs(g1, 0, -1, c1, 0, cnt1)

    t := max(cnt2[0], cnt2[1])
    ans := make([]int, n)
    for i := 0; i < n; i++ {
        ans[i] = t + cnt1[c1[i]]
    }
    return ans
}

func build(edges [][]int) [][]int {
    n := len(edges) + 1
    g := make([][]int, n)
    for _, e := range edges {
        a, b := e[0], e[1]
        g[a] = append(g[a], b)
        g[b] = append(g[b], a)
    }
    return g
}

func dfs(g [][]int, a, fa int, c []int, d int, cnt []int) {
    c[a] = d
    cnt[d]++
    for _, b := range g[a] {
        if b != fa {
            dfs(g, b, a, c, d^1, cnt)
        }
    }
}

# Accepted solution for LeetCode #3373: Maximize the Number of Target Nodes After Connecting Trees II
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)]

// Accepted solution for LeetCode #3373: Maximize the Number of Target Nodes After Connecting Trees II
impl Solution {
    pub fn max_target_nodes(edges1: Vec<Vec<i32>>, edges2: Vec<Vec<i32>>) -> Vec<i32> {
        fn build(edges: &Vec<Vec<i32>>) -> Vec<Vec<i32>> {
            let n = edges.len() + 1;
            let mut g = vec![vec![]; n];
            for e in edges {
                let a = e[0] as usize;
                let b = e[1] as usize;
                g[a].push(b as i32);
                g[b].push(a as i32);
            }
            g
        }

        fn dfs(g: &Vec<Vec<i32>>, a: usize, fa: i32, c: &mut Vec<i32>, d: i32, cnt: &mut Vec<i32>) {
            c[a] = d;
            cnt[d as usize] += 1;
            for &b in &g[a] {
                if b != fa {
                    dfs(g, b as usize, a as i32, c, d ^ 1, cnt);
                }
            }
        }

        let g1 = build(&edges1);
        let g2 = build(&edges2);
        let n = g1.len();
        let m = g2.len();

        let mut c1 = vec![0; n];
        let mut c2 = vec![0; m];
        let mut cnt1 = vec![0; 2];
        let mut cnt2 = vec![0; 2];

        dfs(&g2, 0, -1, &mut c2, 0, &mut cnt2);
        dfs(&g1, 0, -1, &mut c1, 0, &mut cnt1);

        let t = cnt2[0].max(cnt2[1]);
        let mut ans = vec![0; n];
        for i in 0..n {
            ans[i] = t + cnt1[c1[i] as usize];
        }

        ans
    }
}

// Accepted solution for LeetCode #3373: Maximize the Number of Target Nodes After Connecting Trees II
function maxTargetNodes(edges1: number[][], edges2: number[][]): number[] {
    const g1 = build(edges1);
    const g2 = build(edges2);
    const [n, m] = [g1.length, g2.length];
    const c1 = Array(n).fill(0);
    const c2 = Array(m).fill(0);
    const cnt1 = [0, 0];
    const cnt2 = [0, 0];

    dfs(g2, 0, -1, c2, 0, cnt2);
    dfs(g1, 0, -1, c1, 0, cnt1);

    const t = Math.max(...cnt2);
    const ans = Array(n);
    for (let i = 0; i < n; i++) {
        ans[i] = t + cnt1[c1[i]];
    }
    return ans;
}

function build(edges: number[][]): number[][] {
    const n = edges.length + 1;
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    return g;
}

function dfs(g: number[][], a: number, fa: number, c: number[], d: number, cnt: number[]): void {
    c[a] = d;
    cnt[d]++;
    for (const b of g[a]) {
        if (b !== fa) {
            dfs(g, b, a, c, d ^ 1, cnt);
        }
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(n + m)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.