Minimum Time to Collect All Apples in a Tree

Minimum Time to Collect All Apples in a Tree - Problem

Minimum Time to Collect All Apples in a Tree

You're standing at the root of a magical orchard (vertex 0) and need to collect all the precious apples scattered throughout the tree. The orchard is structured as an undirected tree with n vertices numbered from 0 to n-1.

🚶‍♂️ Movement Rules:
• It takes exactly 1 second to walk along any edge
• You must start and end at vertex 0
• You can traverse any edge multiple times

🍎 Goal: Find the minimum time needed to collect all apples and return to the starting position.

Input:
• edges: Array where edges[i] = [a_i, b_i] represents an edge between vertices a_i and b_i
• hasApple: Boolean array where hasApple[i] = true means vertex i has an apple

Output: Integer representing minimum seconds needed to collect all apples

Input & Output

example_1.py — Basic Tree

$ 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

💡 Note: We need to collect apples at nodes 2, 4, and 5. Optimal path: 0→1→4(apple)→1→5(apple)→1→0→2(apple)→0. Total: 1+1+1+1+1+1+1+1 = 8 seconds

example_2.py — Simple Path

$ Input: n = 4, edges = [[0,1],[1,2],[0,3]], hasApple = [true,true,true,true]

› Output: 6

💡 Note: All nodes have apples. Path: 0→1→2(apple)→1→0→3(apple)→0. Since we start at 0 which has an apple, we collect it immediately. Total: 2+2+2 = 6 seconds

example_3.py — No Apples

$ Input: n = 4, edges = [[0,1],[1,2],[0,3]], hasApple = [false,false,false,false]

› Output: 0

💡 Note: No apples to collect, so no travel needed. We stay at the starting position.

Visualization

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Understanding the Visualization

Map the Cave

Build the tree structure to understand all possible paths

Mark Treasure Chambers

Identify which nodes contain apples that need collection

Smart Exploration

Use DFS to explore only passages that lead to treasure

Count Round Trips

Each passage to treasure costs 2 seconds (there and back)

Key Takeaway

🎯 Key Insight: Only traverse edges leading to apple-containing subtrees. Each necessary edge costs exactly 2 seconds (round trip).

Time & Space Complexity

Time Complexity

⏱️

O(k! × n²)

k! permutations of apple nodes, each requiring O(n²) for shortest path calculations

⚠ Quadratic Growth

Space Complexity

O(n)

Space for adjacency list and recursion stack

⚡ Linearithmic Space

Constraints

1 ≤ n ≤ 10⁵
edges.length == n - 1
edges[i].length == 2
0 ≤ a_i, b_i ≤ n - 1
a_i ≠ b_i
hasApple.length == n
The given edges form a valid tree

Asked in

G Google 45 a Amazon 32 f Meta 28 ⊞ Microsoft 22

The optimal solution uses Depth-First Search (DFS) to traverse the tree efficiently. Key insight: we only need to visit edges that lead to subtrees containing apples. Each such edge is traversed exactly twice (going down and coming back up), contributing 2 seconds to the total time. The algorithm runs in O(n) time and uses O(n) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Paths)	O(k! × n²)	O(n)	Generate all possible paths from root, visiting all apple nodes
DFS Optimal Solution	O(n)	O(n)	Use DFS to traverse only subtrees containing apples, count round-trip time

Brute Force (Try All Paths) — Algorithm Steps

Build adjacency list from edges
Find all nodes that have apples
Generate all permutations of apple nodes
For each permutation, calculate shortest path from 0 → apple1 → apple2 → ... → 0
Return minimum total time among all permutations

Visualization

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Step-by-Step Walkthrough

Find Apple Nodes

Identify all vertices that contain apples

Generate Permutations

Create all possible orders to visit apple nodes

Calculate Path Costs

For each permutation, calculate total travel time

Return Minimum

Select the permutation with minimum total time

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>

typedef struct {
    int* data;
    int size;
    int capacity;
} IntArray;

IntArray* createIntArray(int capacity) {
    IntArray* arr = malloc(sizeof(IntArray));
    arr->data = malloc(capacity * sizeof(int));
    arr->size = 0;
    arr->capacity = capacity;
    return arr;
}

void addToArray(IntArray* arr, int value) {
    if (arr->size < arr->capacity) {
        arr->data[arr->size++] = value;
    }
}

int bfs(IntArray** graph, int n, int start, int end) {
    if (start == end) return 0;
    
    int* queue = malloc(n * 2 * sizeof(int));
    bool* visited = calloc(n, sizeof(bool));
    int front = 0, rear = 0;
    
    queue[rear++] = start;
    queue[rear++] = 0;
    visited[start] = true;
    
    while (front < rear) {
        int node = queue[front++];
        int dist = queue[front++];
        
        for (int i = 0; i < graph[node]->size; i++) {
            int neighbor = graph[node]->data[i];
            if (neighbor == end) {
                free(queue);
                free(visited);
                return dist + 1;
            }
            if (!visited[neighbor]) {
                visited[neighbor] = true;
                queue[rear++] = neighbor;
                queue[rear++] = dist + 1;
            }
        }
    }
    
    free(queue);
    free(visited);
    return INT_MAX;
}

int minTimeToCollectApples(int n, int** edges, int edgesSize, bool* hasApple) {
    IntArray** graph = malloc(n * sizeof(IntArray*));
    for (int i = 0; i < n; i++) {
        graph[i] = createIntArray(n);
    }
    
    for (int i = 0; i < edgesSize; i++) {
        addToArray(graph[edges[i][0]], edges[i][1]);
        addToArray(graph[edges[i][1]], edges[i][0]);
    }
    
    int* appleNodes = malloc(n * sizeof(int));
    int appleCount = 0;
    
    for (int i = 0; i < n; i++) {
        if (hasApple[i]) {
            appleNodes[appleCount++] = i;
        }
    }
    
    if (appleCount == 0) {
        free(appleNodes);
        return 0;
    }
    
    // Simple case for small apple count
    int minTime = INT_MAX;
    if (appleCount == 1) {
        minTime = 2 * bfs(graph, n, 0, appleNodes[0]);
    }
    
    free(appleNodes);
    return minTime;
}

int main() {
    int n;
    scanf("%d", &n);
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k! × n²)

k! permutations of apple nodes, each requiring O(n²) for shortest path calculations

⚠ Quadratic Growth

Space Complexity

O(n)

Space for adjacency list and recursion stack

⚡ Linearithmic Space

Constraints

1 ≤ n ≤ 10⁵
edges.length == n - 1
edges[i].length == 2
0 ≤ a_i, b_i ≤ n - 1
a_i ≠ b_i
hasApple.length == n
The given edges form a valid tree

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Paths) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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