0903EasyMathGreedy

Best Handful From the Token Tin

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Trains the technique from

LeetCode 2600K Items With the Maximum Sum

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A tin holds numOnes tokens marked 1, numZeros marked 0 and numNegOnes marked -1.

Take exactly k tokens out of the tin. Return the largest total the marks on those tokens can come to.

Examples

Example 1

Input: numOnes = 7, numZeros = 4, numNegOnes = 6, k = 13
Output: 5

Taking all seven ones and all four zeros accounts for eleven tokens, so the last two must be negatives, giving 7 less 2.

Example 2

Input: numOnes = 3, numZeros = 4, numNegOnes = 0, k = 5
Output: 3

All three ones fit and the remaining two tokens are zeros, so nothing pulls the total down.

Example 3

Input: numOnes = 0, numZeros = 0, numNegOnes = 9, k = 9
Output: -9

The tin holds nothing but negatives and all nine must come out.

Constraints

0 <= numOnes <= 50
0 <= numZeros <= 50
0 <= numNegOnes <= 50
0 <= k <= numOnes + numZeros + numNegOnes

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def k_items_with_maximum_sum(numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int:

Java

public int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k)

The intended approach

Time O(1): a couple of comparisons.Space O(1).

The tokens are worth 1, 0 and -1, so the best handful takes as many ones as it can, fills the rest with zeros, and only then is forced onto the negatives. That gives the total directly: the number of ones taken is `k` or all of them, whichever is smaller, and the negatives taken are however much of `k` is left over once the ones and zeros are exhausted, which is `k` minus their combined count when that is positive. The total is the first count less the second, with no searching at all.

Pythonentry point: k_items_with_maximum_sum

def k_items_with_maximum_sum(numOnes, numZeros, numNegOnes, k):
    ones = numOnes if numOnes < k else k
    left = k - numOnes - numZeros
    negatives = left if left > 0 else 0
    return ones - negatives

The obvious version

Lay the tokens out as an actual list of marks, sort it with the best first, and add up the first `k`. Builds the whole tin rather than reasoning about the counts.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_k_items_with_maximum_sum(numOnes, numZeros, numNegOnes, k):
    tin = []
    for _ in range(numOnes):
        tin.append(1)
    for _ in range(numZeros):
        tin.append(0)
    for _ in range(numNegOnes):
        tin.append(-1)
    tin.sort(reverse=True)
    total = 0
    for i in range(k):
        total += tin[i]
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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