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

Code

Time Complexity: , Space Complexity:

class Solution {
public:
    int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        int res = 0;
        while(k--) {
            if(--numOnes >= 0) res++;
            else if (--numZeros >= 0)  continue;
            else res--;
        }
        return res;
    }
};

Source