Minimum Time to Collect All Apples in a Tree

class Solution {
public:
    int exploreTree(int current, int prev, vector<vector<int>>& graph, vector<bool>& applePresence) {
        int totalCost = 0, timeForChild = 0;
        for (auto childNode : graph[current]) {
            if (childNode == prev) continue;
            timeForChild = exploreTree(childNode, current, graph, applePresence);
            // Only increase total cost if there's an apple in the subtree or at the child node
            if (timeForChild || applePresence[childNode]) totalCost += timeForChild + 2;
        }

        return totalCost;
    }

    int minTime(int n, vector<vector<int>>& edges, vector<bool>& applePresence) {
        vector<vector<int>> graph(n);
        for (auto& edge : edges) {
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
        }
        return exploreTree(0, -1, graph, applePresence);
    }
};

This C++ solution focuses on finding the minimum time required to collect all apples in a tree, represented as an undirected graph. The primary method is structured around a Depth-First Search (DFS) algorithm which starts from the tree root (node 0) and explores child nodes recursively. Here's how the process unfolds:

• Building the Graph: Convert the edge list into an adjacency list. This will represent the tree structure, where each node points to its connected nodes allowing easy traversal.

• DFS Function exploreTree:

  • This function takes the current node, its parent node (to avoid revisiting the parent during recursion), the graph, and the apple presence vector (applePresence) as arguments.
  • As it dives into each child node, it calculates the time required to collect apples from subtrees. This cumulative cost includes 2 units of time for each edge traversed (to and from each child node) if apples are present in the subtree or directly at the child node.
  • The function returns the total time (or cost) for each node's complete subtree, determining if the traversal to a node (and thus the cost of the edge) is necessary based on the presence of apples.

• Main Function minTime:

  • Here, the graph is first constructed using the provided edges.
  • The exploreTree function is called with the root node (0) to initiate the DFS and calculate the overall minimum time to collect all apples starting from the root.

Given that this solution carefully avoids unnecessary traversals by checking apple presence at each step, it optimizes the time calculation efficiently. When implementing this code, ensure that the graph is correctly built and the applePresence accurately represents the presence of apples at each node. The DFS approach guarantees that each path is considered only once, maintaining a manageable time complexity relative to the number of nodes and edges.

class Solution {
    public int calculateTime(int node, int parent, Map<Integer, List<Integer>> graph,
            List<Boolean> applePresent) {
        if (!graph.containsKey(node))
            return 0;

        int totalDuration = 0, subtreeDuration = 0;
        for (int neighbour : graph.get(node)) {
            if (neighbour == parent)
                continue;
            subtreeDuration = calculateTime(neighbour, node, graph, applePresent);
            if (subtreeDuration > 0 || applePresent.get(neighbour))
                totalDuration += subtreeDuration + 2;
        }
        return totalDuration;
    }

    public int minTraversalTime(int nodesCount, int[][] connections, List<Boolean> applePresent) {
        Map<Integer, List<Integer>> adjacencyList = new HashMap<>();
        for (int[] connection : connections) {
            int x = connection[0], y = connection[1];
            adjacencyList.computeIfAbsent(x, k -> new ArrayList<Integer>()).add(y);
            adjacencyList.computeIfAbsent(y, k -> new ArrayList<Integer>()).add(x);
        }
        return calculateTime(0, -1, adjacencyList, applePresent);
    }
}

The provided Java code tackles the problem of determining the minimum time required to collect all apples in a tree, represented as an undirected graph where some nodes contain apples.

The primary approach in the solution involves the following steps:

1. Constructing an adjacency list from the given tree connections, which represents the tree structure where nodes are linked with their direct connections.
2. Implementing a recursive function to calculate the required time to traverse and collect apples starting from an initial node and exploring its neighboring nodes, considering the following:
   • Traverse each child node unless it's the parent node, maintaining tree integrity.
   • Recursively calculate time required for each subtree.
   • Increment the total time if the current child node or any of its subtrees has apples, accounting for time to move to the child and back.
3. Starting the recursive calculation from node 0 and assuming no parent initially (parent as -1).
4. The total traverse time for collecting apples is the sum of times needed to go to and return from each child that leads to subtree containing an apple.

The 'calculateTime' function is key to handling the recursive traversal and time calculation, while 'minTraversalTime' sets up the graph structure and initiates the traversal process.

With these steps outlined, the code efficiently figures out the optimized path to collect all apples with the least amount of backtracking, which is vital for minimizing the overall traversal time. The chosen methodology of using an adjacency list and a recursive DFS traversal strategy effectively addresses the problem's needs.