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:
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 {
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
}
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| 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;
}
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Solution 2#
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| class Solution:
def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
for i, x in enumerate(reward2):
reward1[i] -= x
reward1.sort(reverse=True)
return sum(reward2) + sum(reward1[:k])
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| class Solution {
public int miceAndCheese(int[] reward1, int[] reward2, int k) {
int ans = 0;
int n = reward1.length;
for (int i = 0; i < n; ++i) {
ans += reward2[i];
reward1[i] -= reward2[i];
}
Arrays.sort(reward1);
for (int i = 0; i < k; ++i) {
ans += reward1[n - i - 1];
}
return ans;
}
}
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| class Solution {
public:
int miceAndCheese(vector<int>& reward1, vector<int>& reward2, int k) {
int n = reward1.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += reward2[i];
reward1[i] -= reward2[i];
}
sort(reward1.rbegin(), reward1.rend());
ans += accumulate(reward1.begin(), reward1.begin() + k, 0);
return ans;
}
};
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| func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
for i, x := range reward2 {
ans += x
reward1[i] -= x
}
sort.Ints(reward1)
n := len(reward1)
for i := 0; i < k; i++ {
ans += reward1[n-i-1]
}
return
}
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| function miceAndCheese(reward1: number[], reward2: number[], k: number): number {
const n = reward1.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
ans += reward2[i];
reward1[i] -= reward2[i];
}
reward1.sort((a, b) => b - a);
for (let i = 0; i < k; ++i) {
ans += reward1[i];
}
return ans;
}
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