What is the time complexity of Fruit Into Baskets?

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

What data structures are used to solve Fruit Into Baskets?

This problem uses Array, Hash Table, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Fruit Into Baskets?

Insight 1: The problem can be efficiently solved using a sliding window approach to find the longest continuous subarray containing at most two distinct elements (fruit types). Insight 2: A hash table (or dictionary) is useful to track the frequency of each fruit type within the current window.

Which companies ask LeetCode 904. Fruit Into Baskets in interviews?

This problem is frequently asked at Deutsche Bank, Walmart Labs.

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Fruit Into Baskets - LeetCode 904 Solution

Fruit Into Baskets - Complete Solution Guide

Fruit Into Baskets is LeetCode problem 904, 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

You are visiting a farm that has a single row of fruit trees arranged from left to right. The trees are represented by an integer array fruits where fruits[i] is the type of fruit the i th tree produces. You want to collect as much fruit as possible. However, the owner has some strict rules that you must follow: You only have two baskets, and each basket can only hold a single type of fruit. There is no limit on the amount of fruit each basket can hold. Starting from any tree of your choice, you

Detailed Explanation

The problem "Fruit Into Baskets" asks us to find the maximum number of fruits we can pick from a row of trees, given that we have only two baskets and each basket can hold only one type of fruit. We must start picking from any tree and continue moving right, picking one fruit from each tree, until we encounter a fruit type that doesn't fit in either of our two baskets. The goal is to maximize the number of fruits picked under these constraints.

Solution Approach

The solution uses a sliding window technique combined with a hash map to keep track of the count of each fruit type in the current window. The window expands to the right as long as there are at most two distinct fruit types in it. If the window contains more than two types, we shrink it from the left until the condition is met. In each step, we keep track of the maximum window size observed so far, which represents the maximum number of fruits that can be picked.

Step-by-Step Algorithm

Step 1: Initialize a hash map `basket` to store the frequency of each fruit type within the sliding window, the left pointer of the window `left` to 0, and the maximum picked fruits `max_picked` to 0.
Step 2: Iterate through the `fruits` array using the right pointer `right`. For each fruit encountered, update its count in the `basket` hash map.
Step 3: While the number of distinct fruit types in the `basket` (i.e., the size of the hash map) is greater than 2, shrink the window from the left.
Step 4: To shrink the window, decrement the count of the leftmost fruit type in the `basket`. If its count becomes 0, remove it from the `basket`.
Step 5: Move the left pointer `left` one step to the right.
Step 6: After each iteration of the outer loop, update `max_picked` to be the maximum of its current value and the current window size (right - left + 1).
Step 7: Return `max_picked` after the iteration finishes.

Key Insights

Insight 1: The problem can be efficiently solved using a sliding window approach to find the longest continuous subarray containing at most two distinct elements (fruit types).
Insight 2: A hash table (or dictionary) is useful to track the frequency of each fruit type within the current window.
Insight 3: Maintaining the window size and updating it based on the number of distinct fruit types ensures we meet the problem's constraints.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Sliding Window.

Companies

Asked at: Deutsche Bank, Walmart Labs.

Fruit Into Baskets | LeetCode 904

Medium

Array Hash Table Sliding Window

Time: O(n)Space: O(1)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int totalFruit(int[] fruits) {
        Map<Integer, Integer> basket = new HashMap<>();
        int left = 0;
        int maxPicked = 0;

        for (int right = 0; right < fruits.length; right++) {
            basket.put(fruits[right], basket.getOrDefault(fruits[right], 0) + 1);

            while (basket.size() > 2) {
                int leftFruit = fruits[left];
                basket.put(leftFruit, basket.get(leftFruit) - 1);
                if (basket.get(leftFruit) == 0) {
                    basket.remove(leftFruit);
                }
                left++;
            }

            maxPicked = Math.max(maxPicked, right - left + 1);
        }

        return maxPicked;
    }
}

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Problem Info

Description

Problem 904: Fruit Into Baskets - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window

Companies

Deutsche BankWalmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:904

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

7 steps

1
Step 1: Initialize a hash map `basket` to store the frequency of each fruit type within the sliding window, the left pointer of the window `left` to 0, and the maximum picked fruits `max_picked` to 0.
2
Step 2: Iterate through the `fruits` array using the right pointer `right`. For each fruit encountered, update its count in the `basket` hash map.
3
Step 3: While the number of distinct fruit types in the `basket` (i.e., the size of the hash map) is greater than 2, shrink the window from the left.
4
Step 4: To shrink the window, decrement the count of the leftmost fruit type in the `basket`. If its count becomes 0, remove it from the `basket`.
5
Step 5: Move the left pointer `left` one step to the right.
6
Step 6: After each iteration of the outer loop, update `max_picked` to be the maximum of its current value and the current window size (right - left + 1).
7
Step 7: Return `max_picked` after the iteration finishes.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) due to the single pass through the array using the sliding window. The while loop inside might appear to add complexity, but since `left` can only increment up to `n`, it doesn't change the overall time complexity order. The space complexity is O(1) because the hash table `basket` stores at most three distinct fruit types, regardless of the input size.

📖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

Brute Force

Iterate through all possible subarrays and check if each subarray contains at most two distinct elements. Keep track of the maximum length of such subarrays.

Time: O(n^2), where n is the length of the fruits array, because there are O(n^2) subarrays to consider.Space: O(n) in the worst case, to store the subarray and a hash table to count fruit frequencies, although it could be made O(1) with careful implementation.

Trade-offs: The brute force approach is simple to understand but inefficient due to the nested loops. It is suitable for very small input sizes. The sliding window approach is much more efficient for larger inputs.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using a sliding window approach to find the longest continuous subarray containing at most two distinct elements (fruit types).
2Insight 2: A hash table (or dictionary) is useful to track the frequency of each fruit type within the current window.
3Insight 3: Maintaining the window size and updating it based on the number of distinct fruit types ensures we meet the problem's constraints.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

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

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: Empty fruits array. The algorithm correctly returns 0 as the maximum number of fruits picked.
Edge case 2: Fruits array with only one fruit type. The algorithm correctly returns the length of the array.
Edge case 3: Fruits array with two fruit types. The algorithm correctly returns the length of the array.
Edge case 4: Fruits array where the optimal window is at the end. The algorithm correctly identifies the optimal window and its size.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to shrink the window when the number of distinct fruit types exceeds 2. This leads to an incorrect maximum number of fruits picked.

✗Common mistake 2: Not updating the maximum picked fruits in each iteration. This can lead to missing the optimal solution when the maximum window is encountered somewhere in the middle of the iteration.

✗Common mistake 3: Incorrectly decrementing and removing the fruit count in the `basket` while shrinking the window. If not handled carefully, this can lead to logical errors.

✅

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: Resource allocation where you have limited resources (baskets) and need to maximize the usage of a continuous sequence of tasks or items (fruits).

🏦Real-world scenario 2: Network traffic analysis where you want to find the longest period where only two types of network protocols are being used to optimize data transfer.

🎮Real-world scenario 3: DNA sequencing, where you want to find the longest sequence of DNA that only uses two specific base pairs.

📱Real-world scenario 4: Finding the longest sequence of consecutive days with a maximum of two different weather conditions.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 5, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Fruit Into Baskets"?

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

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