2600. K Items With the Maximum Sum

Problem

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

Solution (Java)

class Solution {
    public int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        int ans[]=new int[numOnes+numZeros+numNegOnes];
        int i=0;
        while(numOnes>0){
            ans[i++]=1;
            numOnes--;
        }
        while(numZeros>0){
            ans[i++]=0;
            numZeros--;
        }

        while(numNegOnes>0){
            ans[i++]=-1;
            numNegOnes--;
        }

        int sum=0;
        int j=0;
        while(k>0){

            sum+=ans[j++];

            k--;
        }
        return sum;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).