0047MediumArrayDynamic Programming

Tower Activation Credits

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Trains the technique from

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 carrier runs a line of relay towers strung along one highway, listed west to east in the integer array credits. Switching on the tower at position i earns credits[i] units of billable bandwidth for the month.

Towers that sit side by side in the line share an overlapping band. Because of that, whenever you switch a tower on, both towers immediately beside it have to stay dark. Any selection obeying that rule is permitted, and leaving every tower dark is permitted too.

Return the greatest total the carrier can bill.

Examples

Example 1

Input: credits = [2, 8, 9, 8, 2]
Output: 16

Lighting the second and fourth towers bills 8 + 8 = 16. Grabbing the fat 9 in the centre forces both eights dark and only reaches 9 + 2 + 2 = 13.

Example 2

Input: credits = [3, 1, 1, 3, 1, 3]
Output: 9

Positions 0, 3 and 5 are pairwise non-neighbouring and bill 3 + 3 + 3 = 9.

Example 3

Input: credits = [7]
Output: 7

A lone tower has no neighbour to interfere with, so it can be lit.

Constraints

1 <= credits.length <= 100
0 <= credits[i] <= 400

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_credits(credits: list[int]) -> int:

Java

public int maxCredits(int[] credits)

The intended approach

Time O(n)Space O(1)

Sweep west to east carrying two running totals: the best billing over everything up to the current tower, and the best billing over everything up to the tower before it. Lighting the current tower can only be added to the second of those, so the new best is the larger of leaving it dark and adding it to the earlier total.

Pythonentry point: max_credits

def max_credits(credits):
    best_here = 0
    best_before = 0
    for value in credits:
        best_here, best_before = max(best_here, best_before + value), best_here
    return best_here

The obvious version

Enumerate every subset of towers as a bit pattern, throw away any pattern with two set bits side by side, and keep the largest surviving total. Exponential but transparently correct.

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_credits(credits):
    n = len(credits)
    best = 0
    for pattern in range(1 << n):
        if pattern & (pattern >> 1):
            continue
        total = 0
        for i in range(n):
            if (pattern >> i) & 1:
                total += credits[i]
        if total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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