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

Master LeetCode problem 2106 (Maximum Fruits Harvested After at Most K Steps), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2106. Maximum Fruits Harvested After at Most K Steps

Hard

Array Binary Search Sliding Window Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will maintain a window of size k starting from the startPos position and keep track of the maximum fruits harvested within this window. We will iteratively expand the window by moving left or right based on which direction gives us more fruits until we have taken k steps.

We will use a hashmap to store the positions of fruits along with the number of fruits at each position for efficient lookup. We will also keep track of the total fruits harvested so far.

Time complexity: O(n) where n is the number of elements in the fruits array
Space complexity: O(n) for the hashmap

Solution Code

import java.util.*;

class Solution {
    public int maxFruits(int[][] fruits, int startPos, int k) {
        Map<Integer, Integer> fruitMap = new HashMap<>();
        int fruitsHarvested = 0;
        int maxFruits = 0;
        int left = startPos;
        int right = startPos;
        
        for (int[] fruit : fruits) {
            fruitMap.put(fruit[0], fruit[1]);
        }
        
        while (k > 0) {
            if (fruitMap.containsKey(left)) {
                fruitsHarvested += fruitMap.get(left);
                maxFruits = Math.max(maxFruits, fruitsHarvested);
                k--;
                left--;
            } else if (fruitMap.containsKey(right)) {
                fruitsHarvested += fruitMap.get(right);
                maxFruits = Math.max(maxFruits, fruitsHarvested);
                k--;
                right++;
            } else {
                if (left == startPos && right == startPos) {
                    break;
                }
                if (left > 0) {
                    left--;
                }
                if (right < 2 * 105) {
                    right++;
                }
            }
        }
        
        return maxFruits;
    }
}

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