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2600. K Items With the Maximum Sum

There is a bag that consists of items, each item has a number 1, 0, or -1 written on it.

You are given four non-negative integers numOnes, numZeros, numNegOnes, and k.

The bag initially contains:

  • numOnes items with 1s written on them.
  • numZeroes items with 0s written on them.
  • numNegOnes items with -1s written on them.

We want to pick exactly k items among the available items. Return the maximum possible sum of numbers written on the items.

Example 1:

Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 2
Output: 2
Explanation: We have a bag of items with numbers written on them {1, 1, 1, 0, 0}. We take 2 items with 1 written on them and get a sum in a total of 2.
It can be proven that 2 is the maximum possible sum.

Example 2:

Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 4
Output: 3
Explanation: We have a bag of items with numbers written on them {1, 1, 1, 0, 0}. We take 3 items with 1 written on them, and 1 item with 0 written on it, and get a sum in a total of 3.
It can be proven that 3 is the maximum possible sum.

Constraints:

  • 0 <= numOnes, numZeros, numNegOnes <= 50
  • 0 <= k <= numOnes + numZeros + numNegOnes

Solutions (Rust)

1. Solution

impl Solution {
    pub fn k_items_with_maximum_sum(
        num_ones: i32,
        num_zeros: i32,
        num_neg_ones: i32,
        k: i32,
    ) -> i32 {
        if num_ones + num_zeros >= k {
            num_ones.min(k)
        } else {
            num_ones * 2 - k + num_zeros
        }
    }
}