LeetCode 3479: Fruits Into Baskets III Solution

Master LeetCode problem 3479 (Fruits Into Baskets III), a medium 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.

3479. Fruits Into Baskets III

Medium

Array Binary Search Segment Tree Ordered Set

Problem Explanation

Explanation

To solve this problem, we can iterate through the fruits and baskets arrays and greedily assign each fruit to a basket with sufficient capacity. We can use a hashmap to keep track of the available baskets for each fruit type. If a fruit cannot be placed in any basket, we increment the count of unplaced fruits. Finally, we return the count of unplaced fruits.

Algorithm:

Initialize a hashmap to store the available baskets for each fruit type.
Iterate through the fruits array:
- For each fruit type, find the first available basket with sufficient capacity.
- If no basket is available, increment the count of unplaced fruits.
Return the count of unplaced fruits.

Time Complexity: O(n) where n is the number of fruits (or baskets)
Space Complexity: O(n)

Solution Code

import java.util.HashMap;

class Solution {
    public int unplacedFruits(int[] fruits, int[] baskets) {
        HashMap<Integer, Integer> basketMap = new HashMap<>();
        int unplacedCount = 0;

        for (int i = 0; i < fruits.length; i++) {
            int fruitType = fruits[i];
            int basketIndex = -1;
            for (int j = 0; j < baskets.length; j++) {
                if (baskets[j] >= fruitType && (!basketMap.containsKey(fruitType) || basketMap.get(fruitType) < j)) {
                    basketIndex = j;
                    break;
                }
            }

            if (basketIndex == -1) {
                unplacedCount++;
            } else {
                basketMap.put(fruitType, basketIndex);
            }
        }

        return unplacedCount;
    }
}

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