2611. Mice and Cheese

Problem

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],&nbsp;reward2[i] <= 1000
• 0 <= k <= n

Solution (Java)

class Solution {
    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
        int n = reward1.length;
        int ans = 0;
        int[] ar = new int[n];
        for(int i = 0; i < n; i++){
            ar[i] = reward1[i]-reward2[i];
            ans += reward2[i];  // adding all of reward2 to ans
        }

        // sorting the array
        Arrays.sort(ar);
        System.out.println(Arrays.toString(ar));

        // Adding the last k because it's sorted in asc order & we need largest elements
        // so k elements from reward2 will be converted to reward1's element
        for(int i = 0; i < k; i++){
            ans += ar[n-1-i];
        }

        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).