LeetCode in Kotlin

2611. Mice and Cheese

Medium

There are two mice and n different types of cheese, each type of cheese should be eaten by exactly one mouse.

A point of the cheese with index i (0-indexed) is:

• reward1[i] if the first mouse eats it.
• reward2[i] if the second mouse eats it.

You are given a positive integer array reward1, a positive integer array reward2, and a non-negative integer k.

Return the maximum points the mice can achieve if the first mouse eats exactly k types of cheese.

Example 1:

Input: reward1 = [1,1,3,4], reward2 = [4,4,1,1], k = 2

Output: 15

Explanation: In this example, the first mouse eats the 2^nd (0-indexed) and the 3^rd types of cheese, and the second mouse eats the 0^th and the 1^st types of cheese. The total points are 4 + 4 + 3 + 4 = 15. It can be proven that 15 is the maximum total points that the mice can achieve.

Example 2:

Input: reward1 = [1,1], reward2 = [1,1], k = 2

Output: 2

Explanation: In this example, the first mouse eats the 0^th (0-indexed) and 1^st types of cheese, and the second mouse does not eat any cheese. The total points are 1 + 1 = 2. It can be proven that 2 is the maximum total points that the mice can achieve.

Constraints:

• 1 <= n == reward1.length == reward2.length <= 105
• 1 <= reward1[i], reward2[i] <= 1000
• 0 <= k <= n

Solution

import java.util.PriorityQueue

class Solution {
    fun miceAndCheese(firstReward: IntArray, seondReward: IntArray, numberOfTypesOfCheeseForFirstMouse: Int): Int {
        var maximumPoints = 0
        val totalTypesOfCheese = firstReward.size
        val minHeapDifferenceInRewards = PriorityQueue<Int>()
        for (i in 0 until totalTypesOfCheese) {
            maximumPoints += firstReward[i]
            minHeapDifferenceInRewards.add(firstReward[i] - seondReward[i])
            if (minHeapDifferenceInRewards.size > numberOfTypesOfCheeseForFirstMouse) {
                maximumPoints -= minHeapDifferenceInRewards.poll()
            }
        }
        return maximumPoints
    }
}

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