What is the time complexity of Maximum Fruits Harvested After at Most K Steps?

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

What data structures are used to solve Maximum Fruits Harvested After at Most K Steps?

This problem uses Array, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Fruits Harvested After at Most K Steps?

Insight 1: The 'fruits' array is sorted, making a sliding window approach viable. Insight 2: We need to efficiently calculate the cost (number of steps) to reach a range of fruits, considering the possible routes: all left, all right, or a combination of left and right, and vice-versa.

Which companies ask LeetCode 2106. Maximum Fruits Harvested After at Most K Steps in interviews?

This problem is frequently asked at Deutsche Bank.

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Maximum Fruits Harvested After at Most K Steps - LeetCode 2106 Solution

Maximum Fruits Harvested After at Most K Steps - Complete Solution Guide

Maximum Fruits Harvested After at Most K Steps is LeetCode problem 2106, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Fruits are available at some positions on an infinite x-axis. You are given a 2D integer array fruits where fruits[i] = [position i , amount i ] depicts amount i fruits at the position position i . fruits is already sorted by position i in ascending order , and each position i is unique . You are also given an integer startPos and an integer k . Initially, you are at the position startPos . From any position, you can either walk to the left or right . It takes one step to move one unit on the x-

Detailed Explanation

The problem asks us to find the maximum number of fruits we can harvest given a sorted list of fruit locations and amounts, a starting position, and a maximum number of steps (k). We can move left or right from our starting position and harvest all fruits at a given position. The goal is to maximize the total fruits harvested within the step limit.

Solution Approach

The solution uses a sliding window approach to explore different ranges of fruits. For each window, the cost (number of steps) to visit all fruits within the window is calculated. If the cost is within the limit (k), the total fruits in that window is considered as a potential maximum. The sliding window expands to the right, and the left boundary shifts right when the cost exceeds k. The maximum fruit count is updated during the sliding process.

Step-by-Step Algorithm

Step 1: Initialize `max_fruits` to 0, `left_ptr` to 0, and `current_sum` to 0.
Step 2: Iterate through the `fruits` array with `right_ptr` from 0 to n-1.
Step 3: Add the amount of fruits at `fruits[right_ptr]` to `current_sum`.
Step 4: While the cost to travel from `fruits[left_ptr]` to `fruits[right_ptr]` exceeds `k`, subtract the amount of fruits at `fruits[left_ptr]` from `current_sum` and increment `left_ptr`.
Step 5: Update `max_fruits` to the maximum of its current value and `current_sum`.
Step 6: Return `max_fruits`.

Key Insights

Insight 1: The 'fruits' array is sorted, making a sliding window approach viable.
Insight 2: We need to efficiently calculate the cost (number of steps) to reach a range of fruits, considering the possible routes: all left, all right, or a combination of left and right, and vice-versa.
Insight 3: Prefix sum can be used for efficient retrieval of the sum of fruits within the sliding window.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: Deutsche Bank.

Maximum Fruits Harvested After at Most K Steps | LeetCode 2106

Hard

Array Binary Search Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int maxTotalFruits(int[][] fruits, int startPos, int k) {
        int n = fruits.length;

        // Helper function to calculate the cost of picking fruits from l_pos to r_pos
        // in relation to the startPos
        
        
        int max_fruits = 0;
        int left_ptr = 0;
        int current_sum = 0;

        for (int right_ptr = 0; right_ptr < n; right_ptr++) {
            current_sum += fruits[right_ptr][1];

            while (left_ptr <= right_ptr && calculate_cost(fruits[left_ptr][0], fruits[right_ptr][0], startPos) > k) {
                current_sum -= fruits[left_ptr][1];
                left_ptr++;
            }

            max_fruits = Math.max(max_fruits, current_sum);
        }

        return max_fruits;
    }

        private int calculate_cost(int l_pos, int r_pos, int startPos) {
            if (l_pos >= startPos) {
                return r_pos - startPos;
            }

            if (r_pos <= startPos) {
                return startPos - l_pos;
            }

            return Math.min(2 * (startPos - l_pos) + (r_pos - startPos),
                    (startPos - l_pos) + 2 * (r_pos - startPos));
        }
}

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Problem Info

Description

Problem 2106: Maximum Fruits Harvested After at Most K Steps - Detailed explanation and solution approaches.

Topics

Array Binary Search Sliding Window Prefix Sum

Companies

Deutsche Bank

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2106

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: Initialize `max_fruits` to 0, `left_ptr` to 0, and `current_sum` to 0.
2
Step 2: Iterate through the `fruits` array with `right_ptr` from 0 to n-1.
3
Step 3: Add the amount of fruits at `fruits[right_ptr]` to `current_sum`.
4
Step 4: While the cost to travel from `fruits[left_ptr]` to `fruits[right_ptr]` exceeds `k`, subtract the amount of fruits at `fruits[left_ptr]` from `current_sum` and increment `left_ptr`.
5
Step 5: Update `max_fruits` to the maximum of its current value and `current_sum`.
6
Step 6: Return `max_fruits`.

💡

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 linear because the `left_ptr` and `right_ptr` each traverse the `fruits` array at most once. The space complexity is constant because the amount of extra space used does not depend on the size of the input.

📖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

Binary Search with Prefix Sum

Create a prefix sum array of fruit amounts. Then for each possible range (left and right boundaries), use binary search to find the boundaries within the allowed steps (k). This allows efficient calculation of total fruit amounts within reachable ranges.

Time: O(n log n), where n is the number of fruits due to the binary search component.Space: O(n) for the prefix sum array.

Trade-offs: This approach can be more efficient than the sliding window if k is very large, making the sliding window cover a significant portion of the fruits array repeatedly. However, the sliding window approach is often simpler to implement and performs well in practice.

✨

Key Insights

1Insight 1: The 'fruits' array is sorted, making a sliding window approach viable.
2Insight 2: We need to efficiently calculate the cost (number of steps) to reach a range of fruits, considering the possible routes: all left, all right, or a combination of left and right, and vice-versa.
3Insight 3: Prefix sum can be used for efficient retrieval of the sum of fruits within the sliding window.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

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

Edge Cases

Edge case 1: No fruits within reach (k is small or fruits are far). The sliding window won't find any valid range, resulting in `max_fruits` remaining 0, which is the correct answer.
Edge case 2: startPos is beyond the range of the fruits. The algorithm correctly adjusts the left and right pointer and only considers the reachable fruits. The `calculate_cost` function handles the position comparisons correctly.
Edge case 3: Large number of fruits close to startPos and k allows harvesting them all. The sliding window expands to include all reachable fruits, leading to maximum harvest.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the cost of traveling to the fruits. The cost needs to consider traveling from startPos to the leftmost fruit, and then the extra distance required to get to the rightmost fruit. One direction should only be traversed once while the other is traversed twice.

✗Common mistake 2: Not handling cases where all fruits are either to the left or right of the starting position. `calculate_cost` must correctly handle those cases where the movement is unidirectional.

✗Common mistake 3: Not initializing variables correctly, leading to wrong result. e.g., set max_fruits to 0 initially

✅

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 with distance constraints. Consider a delivery service with a limited travel range (k) that needs to pick up packages (fruits) at different locations. The algorithm can help determine the maximum number of packages that can be collected within the travel limit.

🏦Real-world scenario 2: Robot harvesting in a farm. A robot needs to harvest crops (fruits) at different locations with a limited battery life (k). The algorithm can optimize the harvesting path to maximize the number of crops harvested before the battery runs out.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: October 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

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.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Maximum Fruits Harvested After at Most K Steps"?

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

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