1443. Minimum Time to Collect All Apples in a Tree

Problem Statement

1443. Minimum Time to Collect All Apples in a Tree

Medium

Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex.

The edges of the undirected tree are given in the array edges, where edges[i] = [a_i, b_i] means that exists an edge connecting the vertices a_i and b_i. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
Output: 8 
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
Output: 6
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.

Example 3:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
Output: 0

Constraints:

1 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
0 <= a_i < b_i <= n - 1
from_i < to_i
hasApple.length == n

C++

Source file

class Solution {
public:
    unordered_map<int, vector<int>> g; // to store the graph
    unordered_map<int, bool> visited; // to stop exploring same nodes again and again.
	
    void createGraph(vector<vector<int>>& edges) {
      for (auto e: edges) {
        g[e[0]].push_back(e[1]); // adjecency list representation
		g[e[1]].push_back(e[0]); // adjecency list representation
      }
    }
  
    int dfs(int node, int myCost, vector<bool>& hasApple) {
	  if (visited[node]) {
		  return 0;
	  }
	  visited[node] = true;
	  
      int childrenCost = 0; // cost of traversing all children. 
      for (auto x: g[node]) { 
        childrenCost += dfs(x, 2, hasApple);  // check recursively for all apples.
      }

      if (childrenCost == 0 && hasApple[node] == false) {
	  // If no child has apples, then we won't traverse the subtree, so cost will be zero.
	  // similarly, if current node also does not have the apple, we won't traverse this branch at all, so cost will be zero.
        return 0;
      }
	  
	  // Children has at least one apple or the current node has an apple, so add those costs.
      return (childrenCost + myCost);
    }
  
    int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
      createGraph(edges); // construct the graph first.
      return dfs(0, 0, hasApple); // cost of reaching the root is 0. For all others, its 2.
    }
};