LeetCode in Kotlin

3477. Fruits Into Baskets II

Easy

You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the ith type of fruit, and baskets[j] represents the capacity of the jth basket.

From left to right, place the fruits according to these rules:

Return the number of fruit types that remain unplaced after all possible allocations are made.

Example 1:

Input: fruits = [4,2,5], baskets = [3,5,4]

Output: 1

Explanation:

Since one fruit type remains unplaced, we return 1.

Example 2:

Input: fruits = [3,6,1], baskets = [6,4,7]

Output: 0

Explanation:

Since all fruits are successfully placed, we return 0.

Constraints:

Solution

class Solution {
    fun numOfUnplacedFruits(fruits: IntArray, baskets: IntArray): Int {
        val n = fruits.size
        var currfruits: Int
        var count = 0
        for (i in 0..<n) {
            currfruits = fruits[i]
            for (j in 0..<n) {
                if (baskets[j] >= currfruits) {
                    count++
                    baskets[j] = 0
                    break
                }
            }
        }
        return n - count
    }
}