# 2611. Mice and Cheese

## Description

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

## Solutions

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

• class Solution {
public:
int miceAndCheese(vector<int>& reward1, vector<int>& reward2, int k) {
int n = reward1.size();
vector<int> idx(n);
iota(idx.begin(), idx.end(), 0);
sort(idx.begin(), idx.end(), [&](int i, int j) { return reward1[j] - reward2[j] < reward1[i] - reward2[i]; });
int ans = 0;
for (int i = 0; i < k; ++i) {
ans += reward1[idx[i]];
}
for (int i = k; i < n; ++i) {
ans += reward2[idx[i]];
}
return ans;
}
};

• class Solution:
def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
n = len(reward1)
idx = sorted(range(n), key=lambda i: reward1[i] - reward2[i], reverse=True)
return sum(reward1[i] for i in idx[:k]) + sum(reward2[i] for i in idx[k:])


• func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
n := len(reward1)
idx := make([]int, n)
for i := range idx {
idx[i] = i
}
sort.Slice(idx, func(i, j int) bool {
i, j = idx[i], idx[j]
return reward1[j]-reward2[j] < reward1[i]-reward2[i]
})
for i := 0; i < k; i++ {
ans += reward1[idx[i]]
}
for i := k; i < n; i++ {
ans += reward2[idx[i]]
}
return
}

• function miceAndCheese(
reward1: number[],
reward2: number[],
k: number,
): number {
const n = reward1.length;
const idx = Array.from({ length: n }, (_, i) => i);
idx.sort((i, j) => reward1[j] - reward2[j] - (reward1[i] - reward2[i]));
let ans = 0;
for (let i = 0; i < k; ++i) {
ans += reward1[idx[i]];
}
for (let i = k; i < n; ++i) {
ans += reward2[idx[i]];
}
return ans;
}