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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

Solution 1: Greedy + Sort

We can first give all the cheese to the second mouse. Next, consider giving $k$ pieces of cheese to the first mouse. How should we choose these $k$ pieces of cheese? Obviously, if we give the $i$-th piece of cheese from the second mouse to the first mouse, the change in the score is $reward1[i] - reward2[i]$. We hope that this change is as large as possible, so that the total score is maximized.

Therefore, we sort the cheese in decreasing order of reward1[i] - reward2[i]. The first $k$ pieces of cheese are eaten by the first mouse, and the remaining cheese is eaten by the second mouse to obtain the maximum score.

Time complexity $O(n \times \log n)$, space complexity $O(n)$. Where $n$ is the number of cheeses.

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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:])

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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;
    }
}

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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;
    }
};

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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 {