K Items With the Maximum Sum

Problem 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.

Examples

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

Approach

To get the maximum sum possible we first need to select number of 1's and then if k is still more then count of 1's then we need to select some 0's which will add 0 to our sum, after selecting 1's and 0's if k is still positive then we need to select -1's which will reduce our sum.

Complexity

Time complexity: $O(1)$ Space complexity: $O(1)$

Solution

Code in Different Languages

C++

class Solution {
public:
   int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
           int total=0;
   		
           int select=min(numOnes,k);       //selecting ones 
   		total+=select;                            //adding to sum
           k-=select;                                  // reducing k
       
           select=min(numZeros,k);          // selecting 0's if k is non zero otherwise select will be 0
           k-=select;                                 // reducing k if k is non zero
       
           select=min(numNegOnes,k);  // selecting -1's if k is non zero otherwise select will be 0
           total-=select;                          //reducing the score by select because we selected select no of -1's in case of k =0 we will reduce 0
       
           return total;
   }
};

JAVA

class Solution {
    public int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
        
        int max = 0; // the maximum sum
        
        // while we still need to select k numbers
        while(k > 0){
            
            // if we have at least one 1 left, add it to the sum
            if(numOnes > 0){
                numOnes--;
                max++;
                k--;
            }
            // if we have at least one 0 left, skip it
            else if(numZeros > 0){
                numZeros--;
                k--;
            }
            // if we have at least one -1 left, add it to the sum
            else{
                numNegOnes--;
                max--;
                k--;
            }   
        } 
        // if we did not select k numbers, return the maximum sum anyway
        return max;
    }
}

PYTHON

class Solution:
    def kItemsWithMaximumSum(self, numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int:
        if numOnes > k:
            return k
        rem = k - numOnes - numZeros
        if rem <= 0:
            return numOnes
        return numOnes - rem
        

Complexity Analysis

• Time Complexity: $O(1)$

• Space Complexity: $O(1)$

References

• LeetCode Problem: K Items With the Maximum Sum