LeetCode 3477: Fruits Into Baskets II Solution

Master LeetCode problem 3477 (Fruits Into Baskets II), a easy 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.

3477. Fruits Into Baskets II

Easy

Array Binary Search Segment Tree Simulation

Problem Explanation

Explanation:

To solve this problem, we can iterate through the fruits array and try to place each fruit in the available baskets according to the given rules. We can keep track of the remaining capacity of each basket and place the fruit in the leftmost available basket with sufficient capacity. If no basket has enough capacity for a fruit type, we count it as unplaced. Finally, we return the count of unplaced fruit types.

Algorithm:

Initialize a variable unplaced to 0 to count the number of unplaced fruit types.
Iterate through the fruits array.
For each fruit, iterate through the baskets array to find the leftmost available basket with sufficient capacity.
If a suitable basket is found, decrement the capacity of that basket and place the fruit.
If no suitable basket is found, increment the unplaced count.
Return the unplaced count as the final result.

Time Complexity: O(n^2) where n is the length of the fruits and baskets arrays. Space Complexity: O(1)

Solution Code

class Solution {
    public int countUnplacedFruits(int[] fruits, int[] baskets) {
        int unplaced = 0;
        for (int fruit : fruits) {
            boolean placed = false;
            for (int i = 0; i < baskets.length; i++) {
                if (baskets[i] >= fruit) {
                    baskets[i] -= fruit;
                    placed = true;
                    break;
                }
            }
            if (!placed) {
                unplaced++;
            }
        }
        return unplaced;
    }
}

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