1443. Minimum Time to Collect All Apples in a Tree


Problem Description

The problem presents an undirected tree with n vertices. Each vertex in the tree is numbered from 0 to n-1 and can contain zero or more apples. You start at vertex 0 and must find the shortest amount of time to collect all the apples in the tree and return to the starting vertex.

Walking from one vertex to another over an edge takes 1 second. The collection of edges forming the tree is given in the edges array, where each element is a pair [a, b] indicating a bidirectional connection between vertices a and b.

A separate boolean array hasApple indicates whether a particular vertex contains an apple. Specifically, hasApple[i] = true means that vertex i has an apple.

The goal is to calculate the minimum time in seconds needed to traverse the tree, collect all the apples, and return to vertex 0. Note that additional time should not be spent traversing to vertices that do not lead to an apple or are not on the path to collect an apple.

Intuition

The intuition behind the solution to this problem involves performing a Depth-First Search (DFS) from the starting vertex. DFS is particularly useful for this scenario as it allows to explore the tree in a path-oriented manner, visiting all vertices necessary to collect the apples.

The key is to eliminate unnecessary movements. If a subtree does not contain any apples, visiting it would be a waste of time. Therefore, while performing DFS, only traverse to children vertices that have an apple or lead to one.

By using DFS, we start at the root (vertex 0) and traverse down each branch, only continuing as long as there are apples in that direction. If we reach a vertex without an apple and none of its children have apples, we do not need to go further down that path.

When moving back up the tree (after visiting all descendants of a vertex), collect the time spent unless it is the root vertex or the branch did not lead to any apples.

The algorithm also keeps track of visited vertices to avoid redundant traversals, as a vertex may connect to multiple others, creating potential cycles that would not exist in a tree structure.

In summary, the DFS traverses down to collect apples and adds the time cost accordingly, while avoiding traversals of paths that do not contribute to the goal of apple collection.

Learn more about Tree, Depth-First Search and Breadth-First Search patterns.

Solution Approach

The solution to our problem uses Depth-First Search (DFS), a common algorithm for traversing or searching tree or graph data structures. The specific implementation begins with the root of the tree, vertex 0, and explores as far as possible along each branch before backtracking.

Here are the main components of the implementation:

  • A graph representation g (using a defaultdict of lists) of the undirected tree that enables easy access to connected vertices.
  • A list vis to keep track of visited vertices, which is necessary for DFS to prevent revisiting the same vertex, causing infinite loops.
  • A recursive dfs function that encapsulates the logic for traversal and time calculation.

The dfs function performs the following:

  1. Accepts a vertex u and a cost parameter (the time taken to travel to this vertex from its parent).
  2. Checks if u has already been visited to prevent re-traversal. If visited, it returns 0 because no additional cost should be incurred.
  3. If not visited, marks u as visited.
  4. Initiates a variable nxt_cost to aggregate the cost of DFS in the subtree rooted at u.
  5. Iterates over all the children v of u (accessible directly via an edge) and adds the result of the recursive DFS for child v to nxt_cost. The cost for moving to any child is 2 seconds (1 second to go and 1 second to possibly return).
  6. After exploring all children, decides whether to include the cost for u. If u has no apple and nxt_cost is 0 (meaning neither u nor any of its descendants have apples), it returns 0. Otherwise, it includes the cost to travel to u (cost) and any costs from its descendants (nxt_cost).

Finally, the root call to dfs for vertex 0 does not need any cost associated with it—it is the starting point, so the initial cost is 0. The total time to collect all the apples is computed by the accumulation of costs from each DFS call.

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Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have n = 5 vertices and the following tree structure with apples:

1Edges: [[0, 1], [0, 2], [1, 3], [1, 4]]
2hasApple: [false, true, false, true, false]

There is an apple at vertex 1 and at vertex 3. Vertex 0 is where we start and do not have any apples.

Following the algorithm's steps:

  1. Start DFS at vertex 0. It doesn't have an apple, but we don't consider the cost here, as it's our starting point.
  2. Look at the children of 0, which are 1 and 2.
  3. Move to vertex 1 (cost is now 2, 1 to go and 1 to potentially return).
  4. At vertex 1, we have an apple, so we will include this cost.
  5. Continue DFS from 1 to its children, which are 3 and 4.
  6. Visit 3 (cost at 3 is 2), there is an apple, so we collect it and return this cost.
  7. Visit 4 (cost at 4 is 2), no apple, and no children with apples, so ignore this path and return 0 cost.
  8. We backtrack to 1 with the total cost from its subtree: 2 (to 3 and back) + 0 (ignoring 4).
  9. Vertex 2 has no apples and no children with apples, so ignore this path.
  10. The total minimum time used is the cost to travel to 1 (2 seconds) and collect the apple, plus the cost to and from 3 (2 seconds). Thus, the total time is 4 seconds to collect all apples and return to 0.

Here are the visualization steps:

1Step: Vertex - Action - Cost Accumulated
20: Start at root
31: 0 -> 1 - Move - Cost: 0
42: Collect apple at 1 - Include Cost: 2 (1 go + 1 return)
53: 1 -> 3 - Move - Cost: 2 (previous cost)
64: Collect apple at 3 - Include Cost: 4 (2 go + 2 return)
75: 3 -> 1 - Return - Cost: 4
86: 1 -> 0 - Return to start - Cost: 4
97: 0 -> 2 - Check path - Cost: 4
108: No apple at 2 and no children, ignore
119: Finish - Total Minimum Time: 4 seconds

This walk through demonstrates how the DFS method efficiently guides movements towards only the necessary vertices, avoiding wasted time on paths without apples.

Solution Implementation

1from collections import defaultdict
2
3class Solution:
4    def minTime(self, n: int, edges: List[List[int]], has_apple: List[bool]) -> int:
5        # Helper function to perform depth-first search from the current node.
6        def dfs(current_node, cost_to_come_back):
7            # If the current node has been visited, no need to do anything.
8            if visited[current_node]:
9                return 0
10            visited[current_node] = True
11          
12            # Accumulated cost of visiting children nodes.
13            total_cost = 0
14            for child_node in graph[current_node]:
15                # Perform DFS on child nodes with the cost of coming back (2 units).
16                total_cost += dfs(child_node, 2)
17          
18            # If there's no apple at the current node, and no cost accumulated from children,
19            # it means we don't need to collect any apples on this path.
20            if not has_apple[current_node] and total_cost == 0:
21                return 0
22          
23            # Return the total cost of picking apples from this subtree (including the cost to come back).
24            return cost_to_come_back + total_cost
25
26        # Construct the graph using adjacency lists.
27        graph = defaultdict(list)
28        for u, v in edges:
29            graph[u].append(v)
30            graph[v].append(u)
31      
32        # Initialize a visited list to avoid revisiting nodes.
33        visited = [False] * n
34      
35        # Perform a DFS starting at node 0 with no initial cost.
36        # Since we start at the root and don't need to return, we pass 0 as the cost.
37        return dfs(0, 0)
38
1class Solution {
2
3    public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
4        // Visited array to keep track of visited nodes
5        boolean[] visited = new boolean[n];
6      
7        // Graph representation using adjacency lists
8        List<Integer>[] graph = new List[n];
9        Arrays.setAll(graph, x -> new ArrayList<>());
10      
11        // Building the undirected graph from the edges
12        for (int[] edge : edges) {
13            int from = edge[0], to = edge[1];
14            graph[from].add(to);
15            graph[to].add(from);
16        }
17      
18        // Starting DFS from node 0 with initial cost as 0
19        return dfs(0, 0, graph, hasApple, visited);
20    }
21
22    private int dfs(int currentNode, int accumulatedCost, List<Integer>[] graph, 
23                    List<Boolean> hasApple, boolean[] visited) {
24        // If the current node is already visited, we return 0 as no additional cost is needed
25        if (visited[currentNode]) {
26            return 0;
27        }
28      
29        // Mark the current node as visited
30        visited[currentNode] = true;
31
32        // Variable to keep track of the total cost from child nodes
33        int childrenCost = 0;
34      
35        // Traverse through all the adjacent nodes
36        for (int adjacentNode : graph[currentNode]) {
37            childrenCost += dfs(adjacentNode, 2, graph, hasApple, visited);
38        }
39      
40        // If the current node does not have an apple and none of its children have one,
41        // we need not take any actions; hence return 0 cost.
42        if (!hasApple.get(currentNode) && childrenCost == 0) {
43            return 0;
44        }
45      
46        // Accumulated cost includes the cost from child nodes and possibly the cost for visiting current node
47        return accumulatedCost + childrenCost;
48    }
49}
50
1class Solution {
2public:
3    // The main function that calculates the minimum time to collect all apples.
4    int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
5        vector<bool> visited(n, false);          // Keep track of visited nodes
6        vector<vector<int>> graph(n);            // Adjacency list representation of the graph
7
8        // Building the undirected graph
9        for (auto& edge : edges) {
10            int u = edge[0], v = edge[1];
11            graph[u].push_back(v);
12            graph[v].push_back(u);
13        }
14
15        // Start DFS from the node 0 (root) with an initial cost of 0
16        return dfs(0, 0, graph, hasApple, visited);
17    }
18
19private:
20    // Depth-first search function to navigate the tree and calculate the cost.
21    int dfs(int node, int cost, vector<vector<int>>& graph, 
22            vector<bool>& hasApple, vector<bool>& visited) {
23
24        // If the node is already visited, skip it to prevent cycles
25        if (visited[node]) return 0;
26        visited[node] = true;    // Mark the current node as visited
27      
28        int totalCost = 0;       // Total cost to collect apples from child nodes
29
30        // Explore all adjacent nodes (children)
31        for (int& child : graph[node]) {
32            // Add to the total cost the cost of collecting apples from the child subtree
33            totalCost += dfs(child, 2, graph, hasApple, visited);
34        }
35
36        // If the node does not have an apple and there is no cost incurred from its children,
37        // it means we don't need to collect apples from this path, hence return 0.
38        if (!hasApple[node] && totalCost == 0) return 0;
39
40        // Otherwise, return the cost of the current path plus the totalCost from children.
41        return cost + totalCost;
42    }
43};
44
1// Define the type for the edges as an array of number arrays.
2type Edges = number[][];
3
4// The main function calculates the minimum time to collect all apples.
5function minTime(n: number, edges: Edges, hasApple: boolean[]): number {
6    const visited: boolean[] = new Array(n).fill(false); // Keep track of visited nodes
7    const graph: number[][] = new Array(n).fill(0).map(() => []); // Adjacency list representation of the graph
8
9    // Building the undirected graph
10    edges.forEach(edge => {
11        const [u, v] = edge;
12        graph[u].push(v);
13        graph[v].push(u);
14    });
15
16    // Start DFS from the node 0 (root) with an initial cost of 0
17    return dfs(0, 0, graph, hasApple, visited);
18}
19
20// Depth-first search function to navigate the tree and calculate the cost.
21function dfs(
22    node: number,
23    cost: number,
24    graph: number[][],
25    hasApple: boolean[],
26    visited: boolean[]
27): number {
28    // If the node is already visited, skip it to prevent cycles
29    if (visited[node]) return 0;
30    visited[node] = true; // Mark the current node as visited
31
32    let totalCost = 0; // Total cost to collect apples from child nodes
33
34    // Explore all adjacent nodes (children)
35    for (const child of graph[node]) {
36        // Add to the total cost the cost of collecting apples from the child subtree
37        totalCost += dfs(child, 2, graph, hasApple, visited);
38    }
39
40    // If the node does not have an apple and there is no cost incurred from its children,
41    // it means we don't need to collect apples from this path, hence return 0.
42    if (!hasApple[node] && totalCost === 0) return 0;
43
44    // Otherwise, return the cost of the current path plus the totalCost from children.
45    return cost + totalCost;
46}
47

Time and Space Complexity

Time Complexity

The given code implements a Depth-First Search (DFS) algorithm. Each node in the tree is traversed exactly once. For each node, the dfs function is called, which goes through all of the connected child nodes (via the edges).

The DFS algorithm visits each node once and examines each edge once. In the worst case, there would be n-1 edges for n nodes in a tree (as it is an undirected acyclic graph, i.e., a tree structure). Therefore, the time complexity of the function is O(n) because every edge is looked at once, and each node is processed in a single pass.

Space Complexity

The space complexity of the code is based on the space used by the recurrence stack during the DFS calls, the vis array used to mark visited nodes, and the adjacency list representation of the graph g.

  1. Recurrence Stack: In the worst case scenario, where the tree is skewed, the DFS could go as deep as n-1 calls, making the space complexity due to the recursion stack O(n).

  2. Visited Array: An array of size n is used to mark visited nodes, which consumes O(n) space.

  3. Adjacency List: The adjacency list g stores all the edges in a dictionary of lists. In the worst case, a binary tree structure would require about 2*(n-1) space (since each edge {u, v} is stored twice—once for u's list and once for v's list). Therefore, g also occupies O(n) space.

Because these all are additive contributions to the total space used, and none of them dominate the others in the O(n) notation, the overall space complexity of the function is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.


Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:
Question 1 out of 10

What's the output of running the following function using input 56?

1KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13    def dfs(path, res):
14        if len(path) == len(digits):
15            res.append(''.join(path))
16            return
17
18        next_number = digits[len(path)]
19        for letter in KEYBOARD[next_number]:
20            path.append(letter)
21            dfs(path, res)
22            path.pop()
23
24    res = []
25    dfs([], res)
26    return res
27
1private static final Map<Character, char[]> KEYBOARD = Map.of(
2    '2', "abc".toCharArray(),
3    '3', "def".toCharArray(),
4    '4', "ghi".toCharArray(),
5    '5', "jkl".toCharArray(),
6    '6', "mno".toCharArray(),
7    '7', "pqrs".toCharArray(),
8    '8', "tuv".toCharArray(),
9    '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13    List<String> res = new ArrayList<>();
14    dfs(new StringBuilder(), res, digits.toCharArray());
15    return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19    if (path.length() == digits.length) {
20        res.add(path.toString());
21        return;
22    }
23    char next_digit = digits[path.length()];
24    for (char letter : KEYBOARD.get(next_digit)) {
25        path.append(letter);
26        dfs(path, res, digits);
27        path.deleteCharAt(path.length() - 1);
28    }
29}
30
1const KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13    let res = [];
14    dfs(digits, [], res);
15    return res;
16}
17
18function dfs(digits, path, res) {
19    if (path.length === digits.length) {
20        res.push(path.join(''));
21        return;
22    }
23    let next_number = digits.charAt(path.length);
24    for (let letter of KEYBOARD[next_number]) {
25        path.push(letter);
26        dfs(digits, path, res);
27        path.pop();
28    }
29}
30

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