LeetCode 2611. Mice and Cheese Solution in Java, C++, Python & More | Explanation + Code

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 × log n), space complexity O(n). Where n is the number of cheeses.

PythonJavaC++GoTypeScript

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:])(code-box)

class Solution { public int miceAndCheese(int[] reward1, int[] reward2, int k) { int n = reward1.length; Integer[] idx = new Integer[n]; Arrays.setAll(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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

Solution 2

PythonJavaC++GoTypeScript

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])(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)