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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 <= 1051 <= reward1[i], reward2[i] <= 10000 <= 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 { 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; }