What is the time complexity of Minimum Time to Collect All Apples in a Tree?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Time to Collect All Apples in a Tree?

This problem uses Hash Table, Tree, Depth-First Search, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Time to Collect All Apples in a Tree?

Insight 1: We only need to traverse edges that lead to a subtree containing at least one apple. If a subtree doesn't have an apple, there's no need to enter it. Insight 2: A Depth-First Search (DFS) approach is well-suited for exploring the tree and determining which subtrees contain apples.

Which companies ask LeetCode 1443. Minimum Time to Collect All Apples in a Tree in interviews?

This problem is frequently asked at Myntra.

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Minimum Time to Collect All Apples in a Tree - LeetCode 1443 Solution

Minimum Time to Collect All Apples in a Tree - Complete Solution Guide

Minimum Time to Collect All Apples in a Tree is LeetCode problem 1443, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 boolea

Detailed Explanation

The problem asks us to find the minimum time required to collect all apples present in a tree, starting from vertex 0 and returning to vertex 0. We're given an undirected tree represented by its edges and a boolean array indicating whether each node has an apple. Moving from one node to an adjacent node takes 1 second. The goal is to minimize the total time spent traversing the tree to collect all apples and return to the starting point.

Solution Approach

The provided code utilizes a Depth-First Search (DFS) algorithm to solve this problem. The core idea is to traverse the tree, and for each node, determine if its subtree contains any apples. If a subtree rooted at a child node contains an apple (either the child itself has an apple or its subtree contains an apple), then the edge connecting the current node to that child needs to be traversed. Since we need to return to the root, we traverse each such edge twice. The DFS function recursively calculates the time required for each subtree. The base case is implicitly when a node has no children or none of the children's subtrees contain apples.

Step-by-Step Algorithm

Step 1: Build an adjacency list representation of the tree from the given edges. This allows for easy traversal of neighbors for each node.
Step 2: Define a recursive DFS function that takes the current node and its parent as input. This prevents revisiting the parent node.
Step 3: Inside the DFS function, iterate through each child of the current node (excluding the parent).
Step 4: Recursively call the DFS function on each child to determine the time required to collect apples in the child's subtree.
Step 5: If the child's subtree contains apples (either the child itself has an apple or the recursive call returned a non-zero time), add the time from the recursive call plus 2 (for traversing the edge to the child and back) to the total time.
Step 6: Return the total time calculated by the DFS function for the current node.

Key Insights

Insight 1: We only need to traverse edges that lead to a subtree containing at least one apple. If a subtree doesn't have an apple, there's no need to enter it.
Insight 2: A Depth-First Search (DFS) approach is well-suited for exploring the tree and determining which subtrees contain apples.
Insight 3: The total time is the sum of the edge traversals required to reach all apples and return to the starting node. Each relevant edge is traversed twice (once to reach the apple, and once to return).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Tree, Depth-First Search, Breadth-First Search.

Companies

Asked at: Myntra.

Minimum Time to Collect All Apples in a Tree | LeetCode 1443

Medium

Hash Table Tree Depth-First Search Breadth-First Search

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        return dfs(0, -1, adj, hasApple);
    }

    private int dfs(int node, int parent, List<List<Integer>> adj, List<Boolean> hasApple) {
        int total_time = 0;
        for (int child : adj.get(node)) {
            if (child == parent) {
                continue;
            }

            int child_subtree_time = dfs(child, node, adj, hasApple);

            if (child_subtree_time > 0 || hasApple.get(child)) {
                total_time += child_subtree_time + 2;
            }
        }

        return total_time;
    }
}

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Problem Info

Description

Problem 1443: Minimum Time to Collect All Apples in a Tree - Detailed explanation and solution approaches.

Topics

Hash Table Tree Depth-First Search Breadth-First Search

Companies

Myntra

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1443

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Build an adjacency list representation of the tree from the given edges. This allows for easy traversal of neighbors for each node.
2
Step 2: Define a recursive DFS function that takes the current node and its parent as input. This prevents revisiting the parent node.
3
Step 3: Inside the DFS function, iterate through each child of the current node (excluding the parent).
4
Step 4: Recursively call the DFS function on each child to determine the time required to collect apples in the child's subtree.
5
Step 5: If the child's subtree contains apples (either the child itself has an apple or the recursive call returned a non-zero time), add the time from the recursive call plus 2 (for traversing the edge to the child and back) to the total time.
6
Step 6: Return the total time calculated by the DFS function for the current node.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each node and each edge is visited at most once during the DFS traversal. The space complexity is O(n) because the adjacency list stores at most 2 * (n - 1) edges, which is O(n), and the recursion stack can have a maximum depth of n in a skewed tree.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

A BFS approach can also be used to solve this problem. Starting from node 0, we can explore the tree level by level. We maintain a queue of nodes to visit and keep track of the distance from the root. We only add a neighbor to the queue if it leads to a subtree with apples.

Time: O(n)Space: O(n)

Trade-offs: While BFS has the same time and space complexity as DFS, DFS is generally preferred for tree traversal problems due to its simplicity in this case. BFS would require extra bookkeeping to track distances and ensure we revisit edges in both directions.

✨

Key Insights

1Insight 1: We only need to traverse edges that lead to a subtree containing at least one apple. If a subtree doesn't have an apple, there's no need to enter it.
2Insight 2: A Depth-First Search (DFS) approach is well-suited for exploring the tree and determining which subtrees contain apples.
3Insight 3: The total time is the sum of the edge traversals required to reach all apples and return to the starting node. Each relevant edge is traversed twice (once to reach the apple, and once to return).

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: The tree has no apples. In this case, the code correctly returns 0, as no edges need to be traversed.
Edge case 2: The root node (node 0) has an apple, and no other nodes have apples. The code correctly returns 0, as we start at the root and don't need to traverse any edges to collect the apple.
Edge case 3: The tree is a single path, and all nodes have apples. The code calculates the correct path length, doubling each edge since it needs to be traversed twice.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the return path to the root node. The time spent needs to be doubled for the traversal back to the root.

✗Common mistake 2: Not excluding the parent node from the children iteration in the DFS, which would lead to infinite recursion and incorrect time calculation.

✗Common mistake 3: Not handling the case where hasApple[child] is true independently of the recursive call's result. This would result in skipping apple at child itself.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing: Finding the minimum cost path to reach specific nodes in a network that have certain resources (like data servers with specific information).

🏦Real-world scenario 2: Delivery optimization: Determining the shortest route for a delivery truck to visit houses with packages, starting from a depot and returning to the depot.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Time to Collect All Apples in a Tree"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?