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Problem solution · Java

Time Taken to Mark All Nodes

Time Taken to Mark All Nodes: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
83 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Time Taken to Mark All Nodes, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 83 lines of Java from the credited upstream file 3241.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 4 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeTime Taken to Mark All Nodes · JavaJava
Use this to learn the idea, then write your own version.
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);    }  }}

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