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 with1
s written on them.numZeroes
items with0
s written on them.numNegOnes
items with-1
s 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.
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.
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.
0 <= numOnes, numZeros, numNegOnes <= 50
0 <= k <= numOnes + numZeros + numNegOnes
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
}
}
}