1155EasyArrayHash TableGreedy

Keeping the Best Run of Different Chips

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Trains the technique from

LeetCode 3487Maximum Unique Subarray Sum After Deletion

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 tray holds chips whose values are chips. Values may repeat and may be below zero.

Discard as many chips as you like, so long as at least one is left. Then choose a run of neighbouring survivors in which no value turns up twice, and add its values up.

Return the largest total that can be reached.

Examples

Example 1

Input: chips = [1, 2, 3, 3, 2, 1]
Output: 6

Discarding the repeats leaves 1, 2 and 3 side by side, and every value above zero counted once comes to 6.

Example 2

Input: chips = [-3, -1, -2]
Output: -1

Every chip is below zero and one has to be kept, so the least bad single chip is the best on offer.

Example 3

Input: chips = [-5, 4, -3, 4, 2]
Output: 6

Discarding the two negatives and one of the two 4 chips leaves 4 and 2 next to each other for 6.

Constraints

1 <= chips.length <= 100
-100 <= chips[i] <= 100

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 max_sum(chips: list[int]) -> int:

Java

public int maxSum(int[] chips)

The intended approach

Time O(n) over the chips.Space O(n) for the values already counted.

Discarding closes the gaps, so any selection of chips at all can be made into a run of neighbours. That means the position of a chip never matters and the question is only which values to keep. When some chip is above zero, keep exactly one copy of every value above zero. Adding a value above zero can only help, keeping a second copy of it is forbidden, and keeping a zero or a value below zero can only leave the total the same or worse. So the answer is the total of the different values above zero. When no chip is above zero, something still has to be kept, so keep the single largest chip, which is the closest to zero. That is the one case where the answer can come out at or below zero.

Pythonentry point: max_sum

def max_sum(chips):
    best = chips[0]
    for value in chips:
        if value > best:
            best = value
    if best <= 0:
        return best
    total = 0
    seen = set()
    for value in chips:
        if value > 0 and value not in seen:
            seen.add(value)
            total += value
    return total

The obvious version

Try every selection of chips as a bit pattern, throw away the ones holding a repeated value, add up what is left and keep the largest total. Costs two to the power of the chip count.

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_max_sum(chips):
    n = len(chips)
    best = None
    for mask in range(1, 1 << n):
        picked = []
        for i in range(n):
            if mask >> i & 1:
                picked.append(chips[i])
        if len(set(picked)) != len(picked):
            continue
        total = 0
        for value in picked:
            total += value
        if best is None or total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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