Description (leetcode)

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 2^nd (0-indexed) and the 3^rd types of cheese, and the second mouse eats the 0^th and the 1^st 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 0^th (0-indexed) and 1^st 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 <= 10^5
  • 1 <= reward1[i], reward2[i] <= 1000
  • 0 <= k <= n

submission

impl Solution {
    pub fn mice_and_cheese(reward1: Vec<i32>, reward2: Vec<i32>, k: i32) -> i32 {
        // one line for fun
        // the basic idea is to sort the rewords based on diff
        reward1.into_iter()
            .zip(reward2.into_iter())
            .map(|(a, b)| (a - b, b))
            .collect::<std::collections::BinaryHeap<_>>()
            .into_sorted_vec()
            .into_iter()
            .rev()
            .enumerate()
            .map(|(idx, (x, y))| if (idx as i32) < k { x + y } else { y })
            .sum::<i32>()
    }
}