2600

2600. K Items With the Maximum Sum (Easy)

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

Related Topics:
Math, Greedy

Similar Questions:

• Maximum Subarray (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/k-items-with-the-maximum-sum
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        if (k <= numOnes) return k;
        k -= numOnes;
        if (k <= numZeros) return numOnes;
        k -= numZeros;
        return numOnes - min(k, numNegOnes);
    }
};