LeetCode 3241: Time Taken to Mark All Nodes Solution

Master LeetCode problem 3241 (Time Taken to Mark All Nodes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3241. Time Taken to Mark All Nodes

Hard

Dynamic Programming Tree Depth-First Search Graph

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first traversal of the tree while keeping track of the time each node gets marked. We can use a recursive function to propagate the marking time to the adjacent nodes based on the rules provided in the problem.

Start by initializing an array times to store the marking times for each node.
Define a recursive function dfs that takes the current node, its parent, and the current time as parameters.
In the dfs function, check if the current node is odd or even, and mark it accordingly based on its neighbors.
Recur for all the adjacent nodes of the current node, passing the appropriate time based on whether the node is odd or even.
Return the maximum time encountered during the traversal.

The time complexity of this approach is O(N), where N is the number of nodes in the tree, as we visit each node exactly once. The space complexity is also O(N) for the recursive stack.

Solution Code

class Solution {
    public int[] timeTakenToMarkAllNodes(int n, int[][] edges) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            graph.get(edge[0]).add(edge[1]);
            graph.get(edge[1]).add(edge[0]);
        }
        
        int[] times = new int[n];
        dfs(0, -1, 0, graph, times);
        return times;
    }
    
    private int dfs(int node, int parent, int time, List<List<Integer>> graph, int[] times) {
        if (node % 2 == 0) {
            times[node] = Math.max(2, time + 2);
        } else {
            times[node] = Math.max(1, time + 1);
        }
        
        for (int neighbor : graph.get(node)) {
            if (neighbor != parent) {
                times[node] = Math.max(times[node], dfs(neighbor, node, times[node], graph, times));
            }
        }
        
        return times[node];
    }
}

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