K Items With the Maximum Sum

We can solve this problem by analyzing the situation carefully. Here’s an approach to get the maximum possible sum of numbers written on the items:

Approach

1. Pick 1’s: As 1’s are positive, we’ll take as many 1’s as possible until we reach the limit k or exhaust numOnes.
2. Pick 0’s if needed: If we still need more items to reach k after taking all possible 1’s, we’ll take 0’s as they don’t decrease the sum.
3. Pick -1’s if needed: If we still need more items after taking all 1’s and 0’s, we’ll take -1’s as we have no other option.

Steps

1. Calculate initial sum: Initialize the sum with the product of the minimum of numOnes and k (this represents the number of 1’s we take) and then decrease k by the same amount.
2. Check for 0’s: If k is greater than 0, it means we still need more items, but taking 0’s won’t change the sum, so we can simply decrease k by the minimum of k and numZeros.
3. Check for -1’s: If k is still greater than 0, it means we need to take -1’s. The sum will decrease by k since we’re taking k items with -1 written on them.
4. Return the final sum.

Code

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class Solution:
    def kItemsWithMaximumSum(self, numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int:
        # Step 1: Calculate initial sum based on 1's
        initial_sum = min(numOnes, k)
        k -= initial_sum

        # Step 2: Check for 0's
        k -= min(numZeros, k)

        # Step 3: Check for -1's
        final_sum = initial_sum - k

        # Step 4: Return the final sum
        return final_sum

Explanation

The given code takes advantage of the fact that we want to maximize the sum. Hence, we prioritize picking the positive numbers (1’s), then 0’s, and finally the negative numbers (-1’s) if needed. The code ensures that exactly k items are picked and computes the maximum possible sum efficiently.