0490MediumArrayGreedyBit ManipulationPrefix Sum

Boosting the Channel Panel

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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 test bench drives a wall panel from a bank of signal generators. Generator i is described by the positive integer codes[i], and the panel lights a channel whenever at least one generator carries that channel, so the panel's reading is the bitwise OR of all the generator codes.

The bench also has a tray of identical doubler modules. Fitting a doubler onto a generator replaces that generator's code with twice its current value; fitting a second doubler onto the same generator doubles it again. You may fit at most boosts doublers in total, spread across the generators however you like, including several on one generator or none at all.

Return the largest panel reading you can obtain.

Examples

Example 1

Input: codes = [12, 9], boosts = 1
Output: 30

Fitting the doubler to the generator coded 9 turns it into 18, and 18 OR 12 is 30.

Example 2

Input: codes = [7], boosts = 3
Output: 56

The lone generator absorbs all three doublers, taking 7 to 14, then 28, then 56.

Example 3

Input: codes = [1, 2, 4, 8], boosts = 1
Output: 23

Doubling the generator coded 8 gives 16, and 16 OR 4 OR 2 OR 1 is 23.

Constraints

1 <= codes.length <= 10^5
1 <= codes[i] <= 10^9
1 <= boosts <= 15

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_panel_reading(codes: list[int], boosts: int) -> int:

Java

public long bestPanelReading(int[] codes, int boosts)

The intended approach

Time O(n)Space O(n)

Splitting the doublers between two generators wastes them: the best reading always comes from piling all `boosts` of them onto one generator and leaving the rest alone. So build a table of the OR of every suffix of the codes, then sweep left to right carrying the OR of the prefix, and at each position combine prefix, the code shifted left by `boosts`, and the suffix starting just after it. Keep the largest.

Pythonentry point: best_panel_reading

def best_panel_reading(codes, boosts):
    # Every doubler is worth stacking on one single generator, so the answer is
    # the best over generators of "this one doubled boosts times, everybody else
    # untouched". Precompute the OR of every suffix, carry the OR of the prefix
    # as the scan goes, and the two halves surround the doubled generator.
    # time O(n), space O(n) for the suffix table
    n = len(codes)
    suffix = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        suffix[i] = suffix[i + 1] | codes[i]
    best = 0
    prefix = 0
    for i in range(n):
        reading = prefix | (codes[i] << boosts) | suffix[i + 1]
        if reading > best:
            best = reading
        prefix |= codes[i]
    return best

The obvious version

Walk an explicit stack over every way of handing out up to `boosts` doublers, generator by generator, accumulating the running OR as each generator's share is fixed. Keeps the largest reading reached once every generator has been assigned.

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_panel_reading(codes, boosts):
    n = len(codes)
    best = 0
    stack = [(0, boosts, 0)]
    while stack:
        idx, left, acc = stack.pop()
        if idx == n:
            if acc > best:
                best = acc
            continue
        for used in range(left + 1):
            stack.append((idx + 1, left - used, acc | (codes[idx] << used)))
    return best

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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