# Solving Leetcode Interviews in Seconds with AI: Time Taken to Mark All Nodes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3241" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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] = [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.   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.  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **BFS from Each Starting Node:** Iterate through each node in the tree. For each node, perform a Breadth-First Search (BFS) to simulate the marking process.
*   **Time Tracking:** Maintain a `times` array to store the marking time for each node, initialized with infinity. During the BFS, update the `times` array when a node is marked for the first time.
*   **Alternating Marking Rule:**  Implement the even/odd marking rule using the current time `t` in the BFS. Odd nodes are marked if an adjacent node was marked at time `t-1`, and even nodes are marked if an adjacent node was marked at time `t-2`.

*   **Runtime Complexity:** O(n^2) due to the nested loops (iterating through each node and then performing BFS from each node).
*   **Storage Complexity:** O(n) for the adjacency list, the queue in BFS, and the `times` array.

	
	# Code
	```python
	from collections import deque

def solve():
    n = int(input())
    edges = []
    for _ in range(n - 1):
        u, v = map(int, input().split())
        edges.append([u, v])

    def bfs(start_node):
        times = [float('inf')] * n
        times[start_node] = 0
        q = deque([(start_node, 0)])  # (node, time)

        while q:
            node, time = q.popleft()
            
            for neighbor in adj[node]:
                if times[neighbor] == float('inf'):
                    if neighbor % 2 == 1:  # Odd node
                        if times[node] == time and time + 1 < times[neighbor]:  # Check adjacent marking time
                           times[neighbor] = time + 1
                           q.append((neighbor, time + 1))

                    else:  # Even node
                        if times[node] == time and time + 2 < times[neighbor]: # Check adjacent marking time
                            times[neighbor] = time + 2
                            q.append((neighbor, time + 2))
        return times

    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    results = []
    for i in range(n):
        results.append(bfs(i))

    final_results = []
    for i in range(n):
        final_results.append(max(results[i]))

    print(*final_results)


#edges = [[0,1],[0,2]]
#edges = [[0,1]]
#edges = [[2,4],[0,1],[2,3],[0,2]]
#n = 3

# Example Usage: Read input as described in the prompt
if __name__ == "__main__":
    # Read number of nodes
    #n = int(input())

    # Read edges
    #edges = []
    #for _ in range(n - 1):
    #    edges.append(list(map(int, input().split())))

    # Call the solver function.  Modify to take arguments as needed from input.
    #solve(n, edges)
    # Below assumes input is given via STDIN

    # Read n and edges
    import sys
    input = sys.stdin.readline
    n = int(input())
    edges = []
    for _ in range(n - 1):
        u, v = map(int, input().split())
        edges.append([u, v])

    # Function call
    solve()
	```