2600. K Items With the Maximum Sum

中文文档

Description

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

Solution 1: Greedy

According to the problem description, we should take as many items marked as $1$ as possible, then take items marked as $0$, and finally take items marked as $-1$.

Thus:

• If the number of items marked as $1$ in the bag is greater than or equal to $k$, we take $k$ items, and the sum of the numbers is $k$.

• If the number of items marked as $1$ is less than $k$, we take $\textit{numOnes}$ items, resulting in a sum of $\textit{numOnes}$. If the number of items marked as $0$ is greater than or equal to $k - \textit{numOnes}$, we take $k - \textit{numOnes}$ more items, keeping the sum at $\textit{numOnes}$.

• Otherwise, we take $k - \textit{numOnes} - \textit{numZeros}$ items from those marked as $-1$, resulting in a sum of $\textit{numOnes} - (k - \textit{numOnes} - \textit{numZeros})$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def kItemsWithMaximumSum(
        self, numOnes: int, numZeros: int, numNegOnes: int, k: int
    ) -> int:
        if numOnes >= k:
            return k
        if numZeros >= k - numOnes:
            return numOnes
        return numOnes - (k - numOnes - numZeros)

Java

class Solution {
    public int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        if (numOnes >= k) {
            return k;
        }
        if (numZeros >= k - numOnes) {
            return numOnes;
        }
        return numOnes - (k - numOnes - numZeros);
    }
}

C++

class Solution {
public:
    int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        if (numOnes >= k) {
            return k;
        }
        if (numZeros >= k - numOnes) {
            return numOnes;
        }
        return numOnes - (k - numOnes - numZeros);
    }
};

Go

func kItemsWithMaximumSum(numOnes int, numZeros int, numNegOnes int, k int) int {
	if numOnes >= k {
		return k
	}
	if numZeros >= k-numOnes {
		return numOnes
	}
	return numOnes - (k - numOnes - numZeros)
}

TypeScript

function kItemsWithMaximumSum(
    numOnes: number,
    numZeros: number,
    numNegOnes: number,
    k: number,
): number {
    if (numOnes >= k) {
        return k;
    }
    if (numZeros >= k - numOnes) {
        return numOnes;
    }
    return numOnes - (k - numOnes - numZeros);
}

Rust

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 > k {
            return k;
        }

        if num_ones + num_zeros > k {
            return num_ones;
        }

        num_ones - (k - num_ones - num_zeros)
    }
}

C#

public class Solution {
    public int KItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        if (numOnes >= k) {
            return k;
        }
        if (numZeros >= k - numOnes) {
            return numOnes;
        }
        return numOnes - (k - numOnes - numZeros);
    }
}