0063MediumArrayDynamic Programming

Best Multiplier Tile Run

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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 board game deals out a row of scoring tiles. Tile i carries the whole number factors[i], which may be negative, zero or positive.

On your turn you pick one unbroken stretch of neighbouring tiles, hold at least one tile, and walk across the whole stretch. You enter with a score of 1 and your score is multiplied by each tile's number as you step onto it. Skipping a tile inside the stretch is not allowed.

Report the highest score any single stretch can leave you with.

Examples

Example 1

Input: factors = [3, -1, -4, 2]
Output: 24

Walking all four tiles pairs the two negative numbers off against each other, giving 3 * -1 * -4 * 2 = 24. Stopping earlier never gets past 12.

Example 2

Input: factors = [-7, 0, -5]
Output: 0

Both single negative tiles leave a negative score, so the best available stretch is one that steps on the zero tile.

Example 3

Input: factors = [-6]
Output: -6

A stretch has to hold at least one tile, so the only choice available scores -6.

Constraints

1 <= factors.length <= 2 * 10^4
-10 <= factors[i] <= 10
The score of any unbroken stretch of factors is guaranteed to fit in a signed 32-bit integer.

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 best_tile_run(factors: list[int]) -> int:

Java

public int bestTileRun(int[] factors)

The intended approach

Time O(n)Space O(1)

A negative tile turns the worst score so far into the best one, so carry both: the highest and the lowest score of a stretch ending at the current tile. On a negative tile swap the two before extending, then each is either the tile on its own, which restarts the stretch, or the carried score multiplied by the tile. Tracking the largest high seen handles zeros for free, because a zero tile forces both carried scores to restart.

Pythonentry point: best_tile_run

def best_tile_run(factors):
    best = factors[0]
    high = factors[0]
    low = factors[0]
    for value in factors[1:]:
        if value < 0:
            high, low = low, high
        if value > high * value:
            high = value
        else:
            high = high * value
        if value < low * value:
            low = value
        else:
            low = low * value
        if high > best:
            best = high
    return best

The obvious version

Anchor a stretch at every tile, extend it one tile at a time while multiplying, and keep the largest score seen. O(n^2).

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_best_tile_run(factors):
    n = len(factors)
    best = factors[0]
    for start in range(n):
        score = 1
        for end in range(start, n):
            score *= factors[end]
            if score > best:
                best = score
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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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