LeetCode #3241 — HARD

Time Taken to Mark All Nodes

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

The Problem

Problem Statement

There exists an undirected tree with n nodes numbered 0 to n - 1. You are given a 2D integer array edges of length n - 1, where edges[i] = [u_i, v_i] indicates that there is an edge between nodes u_i and v_i in the tree.

Initially, all nodes are unmarked. For each node i:

If i is odd, the node will get marked at time x if there is at least one node adjacent to it which was marked at time x - 1.
If i is even, the node will get marked at time x if there is at least one node adjacent to it which was marked at time x - 2.

Return an array times where times[i] is the time when all nodes get marked in the tree, if you mark node i at time t = 0.

Note that the answer for each times[i] is independent, i.e. when you mark node i all other nodes are unmarked.

Example 1:

Input: edges = [[0,1],[0,2]]

Output: [2,4,3]

Explanation:

For i = 0:
- Node 1 is marked at t = 1, and Node 2 at t = 2.
For i = 1:
- Node 0 is marked at t = 2, and Node 2 at t = 4.
For i = 2:
- Node 0 is marked at t = 2, and Node 1 at t = 3.

Example 2:

Input: edges = [[0,1]]

Output: [1,2]

Explanation:

For i = 0:
- Node 1 is marked at t = 1.
For i = 1:
- Node 0 is marked at t = 2.

Example 3:

Input: edges = [[2,4],[0,1],[2,3],[0,2]]

Output: [4,6,3,5,5]

Explanation:

Constraints:

2 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
0 <= edges[i][0], edges[i][1] <= n - 1
The input is generated such that edges represents a valid tree.

Step 01

Brute Force Baseline

Problem summary: There exists an undirected tree with n nodes numbered 0 to n - 1. You are given a 2D integer array edges of length n - 1, where edges[i] = [ui, vi] indicates that there is an edge between nodes ui and vi in the tree. Initially, all nodes are unmarked. For each node i: If i is odd, the node will get marked at time x if there is at least one node adjacent to it which was marked at time x - 1. If i is even, the node will get marked at time x if there is at least one node adjacent to it which was marked at time x - 2. Return an array times where times[i] is the time when all nodes get marked in the tree, if you mark node i at time t = 0. Note that the answer for each times[i] is independent, i.e. when you mark node i all other nodes are unmarked.

Baseline thinking

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

Pattern signal: Dynamic Programming · Tree

Example 1

[[0,1],[0,2]]

Example 2

[[0,1]]

Example 3

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

Core Insight

What unlocks the optimal approach

Can we use dp on trees?
Store the two most distant children for each node.
When re-rooting the tree, keep a variable for distance to the root node.

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 #3241: Time Taken to Mark All Nodes
class Solution {
  public int[] timeTaken(int[][] edges) {
    final int n = edges.length + 1;
    int[] ans = new int[n];
    List<Integer>[] tree = new List[n];
    // dp[i] := the top two direct child nodes for subtree rooted at node i,
    // where each node contains the time taken to mark the entire subtree rooted
    // at the node itself
    Top2[] dp = new Top2[n];

    for (int i = 0; i < n; ++i) {
      tree[i] = new ArrayList<>();
      dp[i] = new Top2();
    }

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      tree[u].add(v);
      tree[v].add(u);
    }

    dfs(tree, 0, /*prev=*/-1, dp);
    reroot(tree, 0, /*prev=*/-1, /*maxTime=*/0, dp, ans);
    return ans;
  }

  private record Node(int node, int time) {
    Node() {
      this(0, 0);
    }
  }

  private record Top2(Node max1, Node max2) {
    Top2() {
      this(new Node(), new Node());
    }
  }

  // Return the time taken to mark node u.
  private int getTime(int u) {
    return u % 2 == 0 ? 2 : 1;
  }

  // Performs a DFS traversal of the subtree rooted at node `u`, computes the
  // time taken to mark all nodes in the subtree, records the top two direct
  // child nodes, where the time taken to mark the subtree rooted at each of the
  // child nodes is maximized, and returns the top child node.
  //
  // These values are used later in the rerooting process.
  private int dfs(List<Integer>[] tree, int u, int prev, Top2[] dp) {
    Node max1 = new Node();
    Node max2 = new Node();
    for (final int v : tree[u]) {
      if (v == prev)
        continue;
      final int time = dfs(tree, v, u, dp) + getTime(v);
      if (time >= max1.time()) {
        max2 = max1;
        max1 = new Node(v, time);
      } else if (time > max2.time()) {
        max2 = new Node(v, time);
      }
    }
    dp[u] = new Top2(max1, max2);
    return max1.time();
  }

  // Reroots the tree at node `u` and updates the answer array, where `maxTime`
  // is the longest path that doesn't go through `u`'s subtree.
  private void reroot(List<Integer>[] tree, int u, int prev, int maxTime, Top2[] dp, int[] ans) {
    ans[u] = Math.max(maxTime, dp[u].max1().time());
    for (final int v : tree[u]) {
      if (v == prev)
        continue;
      final int newMaxTime =
          getTime(u) +
          Math.max(maxTime, dp[u].max1().node() == v ? dp[u].max2().time() : dp[u].max1().time());
      reroot(tree, v, u, newMaxTime, dp, ans);
    }
  }
}

// Accepted solution for LeetCode #3241: Time Taken to Mark All Nodes
package main

// https://space.bilibili.com/206214
func timeTaken(edges [][]int) []int {
	n := len(edges) + 1
	g := make([][]int, n)
	for _, e := range edges {
		x, y := e[0], e[1]
		g[x] = append(g[x], y)
		g[y] = append(g[y], x)
	}

	// subRes[x] 保存子树 x 的最大深度 maxD，次大深度 maxD2，以及最大深度要往儿子 y 走
	subRes := make([]struct{ maxD, maxD2, y int }, n)
	// 计算 subRes[x]
	var dfs func(int, int)
	dfs = func(x, fa int) {
		res := &subRes[x]
		for _, y := range g[x] {
			if y == fa {
				continue
			}
			dfs(y, x)
			w := 2 - y%2 // 从 x 到 y 的边权
			maxD := subRes[y].maxD + w // 从 x 出发，往 y 方向的最大深度
			if maxD > res.maxD {
				res.maxD2 = res.maxD
				res.maxD = maxD
				res.y = y
			} else if maxD > res.maxD2 {
				res.maxD2 = maxD
			}
		}
	}
	dfs(0, -1)

	// ans[x] 表示当 x 是树根时，整棵树的最大深度
	ans := make([]int, n)
	// 计算 ans[x]
	var reroot func(int, int, int)
	reroot = func(x, fa, fromUp int) {
		sub := subRes[x]
		ans[x] = max(sub.maxD, fromUp)
		for _, y := range g[x] {
			if y == fa {
				continue
			}
			// 站在 x 的角度，不往 y 走，能走多远？
			// 要么往上走（fromUp），要么往除了 y 的其余子树走（mx），二者取最大值
			mx := sub.maxD
			if y == sub.y { // 对于 y 来说，上面要选次大的
				mx = sub.maxD2
			}
			w := 2 - x%2 // 从 y 到 x 的边权
			reroot(y, x, max(fromUp, mx)+w) // 对于 y 来说，加上从 y 到 x 的边权
		}
	}
	reroot(0, -1, 0)
	return ans
}

# Accepted solution for LeetCode #3241: Time Taken to Mark All Nodes
from dataclasses import dataclass


@dataclass
class Node:
  node: int = 0  # the node number
  time: int = 0  # the time taken to mark the entire subtree rooted at the node


class Top2:
  def __init__(self, top1: Node = Node(), top2: Node = Node()):
    # the direct child node, where the time taken to mark the entire subtree
    # rooted at the node is the maximum
    self.top1 = top1
    # the direct child node, where the time taken to mark the entire subtree
    # rooted at the node is the second maximum
    self.top2 = top2


class Solution:
  def timeTaken(self, edges: list[list[int]]) -> list[int]:
    n = len(edges) + 1
    ans = [0] * n
    tree = [[] for _ in range(n)]
    # dp[i] := the top two direct child nodes for subtree rooted at node i,
    # where each node contains the time taken to mark the entire subtree rooted
    # at the node itself
    dp = [Top2()] * n

    for u, v in edges:
      tree[u].append(v)
      tree[v].append(u)

    self._dfs(tree, 0, -1, dp)
    self._reroot(tree, 0, -1, 0, dp, ans)
    return ans

  def _getTime(self, u: int) -> int:
    """Returns the time taken to mark node u."""
    return 2 if u % 2 == 0 else 1

  def _dfs(
      self,
      tree: list[list[int]],
      u: int,
      prev: int,
      dp: list[Top2]
  ) -> int:
    """
    Performs a DFS traversal of the subtree rooted at node `u`, computes the
    time taken to mark all nodes in the subtree, records the top two direct
    child nodes, where the time taken to mark the subtree rooted at each of the
    child nodes is maximized, and returns the top child node.

    These values are used later in the rerooting process.
    """
    top1 = Node()
    top2 = Node()
    for v in tree[u]:
      if v == prev:
        continue
      time = self._dfs(tree, v, u, dp) + self._getTime(v)
      if time >= top1.time:
        top2 = top1
        top1 = Node(v, time)
      elif time > top2.time:
        top2 = Node(v, time)
    dp[u] = Top2(top1, top2)
    return top1.time

  def _reroot(
      self,
      tree: list[list[int]],
      u: int,
      prev: int,
      maxTime: int,
      dp: list[Top2],
      ans: list[int]
  ) -> None:
    """
    Reroots the tree at node `u` and updates the answer array, where `maxTime`
    is the longest path that doesn't go through `u`'s subtree.
    """
    ans[u] = max(maxTime, dp[u].top1.time)

    for v in tree[u]:
      if v == prev:
        continue
      newMaxTime = self._getTime(u) + max(
          maxTime,
          dp[u].top2.time if dp[u].top1.node == v else dp[u].top1.time
      )
      self._reroot(tree, v, u, newMaxTime, dp, ans)

// Accepted solution for LeetCode #3241: Time Taken to Mark All Nodes
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3241: Time Taken to Mark All Nodes
// class Solution {
//   public int[] timeTaken(int[][] edges) {
//     final int n = edges.length + 1;
//     int[] ans = new int[n];
//     List<Integer>[] tree = new List[n];
//     // dp[i] := the top two direct child nodes for subtree rooted at node i,
//     // where each node contains the time taken to mark the entire subtree rooted
//     // at the node itself
//     Top2[] dp = new Top2[n];
// 
//     for (int i = 0; i < n; ++i) {
//       tree[i] = new ArrayList<>();
//       dp[i] = new Top2();
//     }
// 
//     for (int[] edge : edges) {
//       final int u = edge[0];
//       final int v = edge[1];
//       tree[u].add(v);
//       tree[v].add(u);
//     }
// 
//     dfs(tree, 0, /*prev=*/-1, dp);
//     reroot(tree, 0, /*prev=*/-1, /*maxTime=*/0, dp, ans);
//     return ans;
//   }
// 
//   private record Node(int node, int time) {
//     Node() {
//       this(0, 0);
//     }
//   }
// 
//   private record Top2(Node max1, Node max2) {
//     Top2() {
//       this(new Node(), new Node());
//     }
//   }
// 
//   // Return the time taken to mark node u.
//   private int getTime(int u) {
//     return u % 2 == 0 ? 2 : 1;
//   }
// 
//   // Performs a DFS traversal of the subtree rooted at node `u`, computes the
//   // time taken to mark all nodes in the subtree, records the top two direct
//   // child nodes, where the time taken to mark the subtree rooted at each of the
//   // child nodes is maximized, and returns the top child node.
//   //
//   // These values are used later in the rerooting process.
//   private int dfs(List<Integer>[] tree, int u, int prev, Top2[] dp) {
//     Node max1 = new Node();
//     Node max2 = new Node();
//     for (final int v : tree[u]) {
//       if (v == prev)
//         continue;
//       final int time = dfs(tree, v, u, dp) + getTime(v);
//       if (time >= max1.time()) {
//         max2 = max1;
//         max1 = new Node(v, time);
//       } else if (time > max2.time()) {
//         max2 = new Node(v, time);
//       }
//     }
//     dp[u] = new Top2(max1, max2);
//     return max1.time();
//   }
// 
//   // Reroots the tree at node `u` and updates the answer array, where `maxTime`
//   // is the longest path that doesn't go through `u`'s subtree.
//   private void reroot(List<Integer>[] tree, int u, int prev, int maxTime, Top2[] dp, int[] ans) {
//     ans[u] = Math.max(maxTime, dp[u].max1().time());
//     for (final int v : tree[u]) {
//       if (v == prev)
//         continue;
//       final int newMaxTime =
//           getTime(u) +
//           Math.max(maxTime, dp[u].max1().node() == v ? dp[u].max2().time() : dp[u].max1().time());
//       reroot(tree, v, u, newMaxTime, dp, ans);
//     }
//   }
// }

// Accepted solution for LeetCode #3241: Time Taken to Mark All Nodes
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3241: Time Taken to Mark All Nodes
// class Solution {
//   public int[] timeTaken(int[][] edges) {
//     final int n = edges.length + 1;
//     int[] ans = new int[n];
//     List<Integer>[] tree = new List[n];
//     // dp[i] := the top two direct child nodes for subtree rooted at node i,
//     // where each node contains the time taken to mark the entire subtree rooted
//     // at the node itself
//     Top2[] dp = new Top2[n];
// 
//     for (int i = 0; i < n; ++i) {
//       tree[i] = new ArrayList<>();
//       dp[i] = new Top2();
//     }
// 
//     for (int[] edge : edges) {
//       final int u = edge[0];
//       final int v = edge[1];
//       tree[u].add(v);
//       tree[v].add(u);
//     }
// 
//     dfs(tree, 0, /*prev=*/-1, dp);
//     reroot(tree, 0, /*prev=*/-1, /*maxTime=*/0, dp, ans);
//     return ans;
//   }
// 
//   private record Node(int node, int time) {
//     Node() {
//       this(0, 0);
//     }
//   }
// 
//   private record Top2(Node max1, Node max2) {
//     Top2() {
//       this(new Node(), new Node());
//     }
//   }
// 
//   // Return the time taken to mark node u.
//   private int getTime(int u) {
//     return u % 2 == 0 ? 2 : 1;
//   }
// 
//   // Performs a DFS traversal of the subtree rooted at node `u`, computes the
//   // time taken to mark all nodes in the subtree, records the top two direct
//   // child nodes, where the time taken to mark the subtree rooted at each of the
//   // child nodes is maximized, and returns the top child node.
//   //
//   // These values are used later in the rerooting process.
//   private int dfs(List<Integer>[] tree, int u, int prev, Top2[] dp) {
//     Node max1 = new Node();
//     Node max2 = new Node();
//     for (final int v : tree[u]) {
//       if (v == prev)
//         continue;
//       final int time = dfs(tree, v, u, dp) + getTime(v);
//       if (time >= max1.time()) {
//         max2 = max1;
//         max1 = new Node(v, time);
//       } else if (time > max2.time()) {
//         max2 = new Node(v, time);
//       }
//     }
//     dp[u] = new Top2(max1, max2);
//     return max1.time();
//   }
// 
//   // Reroots the tree at node `u` and updates the answer array, where `maxTime`
//   // is the longest path that doesn't go through `u`'s subtree.
//   private void reroot(List<Integer>[] tree, int u, int prev, int maxTime, Top2[] dp, int[] ans) {
//     ans[u] = Math.max(maxTime, dp[u].max1().time());
//     for (final int v : tree[u]) {
//       if (v == prev)
//         continue;
//       final int newMaxTime =
//           getTime(u) +
//           Math.max(maxTime, dp[u].max1().node() == v ? dp[u].max2().time() : dp[u].max1().time());
//       reroot(tree, v, u, newMaxTime, dp, ans);
//     }
//   }
// }

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

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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.

Time Taken to Mark All Nodes

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

State misses one required dimension

Forgetting null/base-case handling