How do you solve Time Taken to Mark All Nodes on LeetCode?

Time Taken to Mark All Nodes (LeetCode #3241) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Marking time depends on the parity of nodes and their adjacency.

What is the brute force solution for Time Taken to Mark All Nodes?

The brute force approach for Time Taken to Mark All Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. For each node, we simulate the marking process by performing a breadth-first search (BFS) or depth-first search (DFS) to determine how long it takes to mark all other nodes. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Time Taken to Mark All Nodes?

Time Taken to Mark All Nodes uses the following concepts: Dynamic Programming, Tree, Depth-First Search, Graph Theory. The recommended patterns to study are: Depth-First Search, Graph Traversal, Dynamic Programming on Trees.

What are the interview tips for Time Taken to Mark All Nodes?

Clarify the problem and constraints before jumping into coding. Think about edge cases, like minimum and maximum node counts. Practice explaining your thought process as you code.

What are common mistakes when solving Time Taken to Mark All Nodes?

Not considering the parity of the nodes correctly. Forgetting to handle the parent-child relationship in DFS.

#3241

Time Taken to Mark All Nodes

Hard

Dynamic ProgrammingTreeDepth-First SearchGraph TheoryDepth-First SearchGraph TraversalDynamic Programming on Trees

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can optimize the marking process using a depth-first search (DFS) approach to calculate the maximum time for marking all nodes based on their distances from the starting node. This allows us to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Build the adjacency list representation of the tree from the edges.
2Step 2: Use DFS to calculate the maximum marking time for each node based on its children.
3Step 3: For each node, determine the time taken to mark all nodes based on the maximum times of its children.

solution.py26 lines

1from collections import defaultdict
2
3def mark_nodes_optimal(edges):
4    n = len(edges) + 1
5    graph = defaultdict(list)
6    for u, v in edges:
7        graph[u].append(v)
8        graph[v].append(u)
9    times = [0] * n
10
11    def dfs(node, parent):
12        max_time_odd = 0
13        max_time_even = 0
14        for neighbor in graph[node]:
15            if neighbor == parent:
16                continue
17            dfs(neighbor, node)
18            if neighbor % 2 == 1:
19                max_time_odd = max(max_time_odd, times[neighbor] + 1)
20            else:
21                max_time_even = max(max_time_even, times[neighbor] + 2)
22        times[node] = max(max_time_odd, max_time_even)
23
24    for start in range(n):
25        dfs(start, -1)
26    return times

ℹ

Complexity note: This complexity is due to the fact that we traverse each node and edge once, leading to O(n) time. The space complexity is O(n) for storing the graph.

1Marking time depends on the parity of nodes and their adjacency.
2Using DFS allows us to efficiently calculate marking times without redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.