LeetCode 2611. Mice and Cheese

原题链接在这里：https://leetcode.com/problems/mice-and-cheese/description/

题目：

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 2nd (0-indexed) and the 3rd types of cheese, and the second mouse eats the 0th and the 1st 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 0th (0-indexed) and 1st 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

题解：

First mouse takes a[i], and second mouse takes b[j], where i and j are in exclusive in the set.

To maximize this, update a[i] + b[j] -> a[i] - b[i] + b[i] + b[j]. 

b[i] + b[j] = sum of b. 

Then it becomes to maximize a[i] - b[i] for k indices.

Use a priority queue for the minimum heap, the size is k.

Time Complexity: O(nlogk). n = reward1.length.

Space: O(k).

AC Java:

class Solution {
    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
        int res = 0;
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        int n = reward1.length;
        for(int i = 0; i < n; i++){
            res += reward2[i];
            minHeap.add(reward1[i] - reward2[i]);
            if(minHeap.size() > k){
                minHeap.poll();
            }
        }

        while(!minHeap.isEmpty()){
            res += minHeap.poll();
        }

        return res;
    }
}