LeetCode in Kotlin

3074. Apple Redistribution into Boxes

Easy

You are given an array apple of size n and an array capacity of size m.

There are n packs where the ith pack contains apple[i] apples. There are m boxes as well, and the ith box has a capacity of capacity[i] apples.

Return the minimum number of boxes you need to select to redistribute these n packs of apples into boxes.

Note that, apples from the same pack can be distributed into different boxes.

Example 1:

Input: apple = [1,3,2], capacity = [4,3,1,5,2]

Output: 2

Explanation: We will use boxes with capacities 4 and 5. It is possible to distribute the apples as the total capacity is greater than or equal to the total number of apples.

Example 2:

Input: apple = [5,5,5], capacity = [2,4,2,7]

Output: 4

Explanation: We will need to use all the boxes.

Constraints:

• 1 <= n == apple.length <= 50
• 1 <= m == capacity.length <= 50
• 1 <= apple[i], capacity[i] <= 50
• The input is generated such that it’s possible to redistribute packs of apples into boxes.

Solution

import kotlin.math.max

class Solution {
    fun minimumBoxes(apple: IntArray, capacity: IntArray): Int {
        val count = IntArray(51)
        var appleSum = 0
        for (j in apple) {
            appleSum += j
        }
        var reqCapacity = 0
        var max = 0
        for (j in capacity) {
            count[j]++
            max = max(max.toDouble(), j.toDouble()).toInt()
        }
        for (i in max downTo 0) {
            if (count[i] >= 1) {
                while (count[i] != 0) {
                    appleSum -= i
                    reqCapacity++
                    if (appleSum <= 0) {
                        return reqCapacity
                    }
                    count[i]--
                }
            }
        }
        return reqCapacity
    }
}

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