# 2611. Mice and Cheese
###### tags: `Leetcode` `Medium`
Link: https://leetcode.com/problems/mice-and-cheese/description/
## 思路
Assume take all from the second array.
Check the difference sequence:
A[0] - B[0], A[1] - B[1], ...

Take k largest from the sequence and sum up.

Return the res = sum(B) + sum(k largest A[i]-B[i])

Python:O(klogk)
Java:O(nlogn)
## Code

class Solution {
    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
        int ans = 0;
        int n = reward1.length;
        int[] diff = new int[n];
        for(int i=0; i<n; i++){
            diff[i] = reward1[i]-reward2[i];
            ans += reward2[i];
        }
        Arrays.sort(diff);
        for(int i=0; i<k; i++){
            ans += diff[n-i-1];
        }
        return ans;
    }
}

class Solution:
    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
        return sum(reward2)+sum(heapq.nlargest(k, (a-b for a,b in zip(reward1, reward2))))