0090MediumArrayBacktrackingBit Manipulation

Every Gain Preset

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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 mixing desk exposes a bank of gain trims, given as the integer array offsets. Every trim in the bank carries a different value, and a value may be negative (a cut) or positive (a boost).

A preset is any selection of trims to engage. Engaging none of them is a preset, and so is engaging the whole bank. Build the catalogue of every preset the desk can hold.

Each preset must list its engaged trims in the same relative order they occupy in offsets. The catalogue itself may come back in any order.

Examples

Example 1

Input: offsets = [4, -2]
Output: [[], [4], [-2], [4, -2]]

Two trims give four presets: nothing engaged, each one alone, and both together with 4 still listed ahead of -2.

Example 2

Input: offsets = [7]
Output: [[], [7]]

A single trim is either idle or engaged.

Example 3

Input: offsets = [0, 3, -5]
Output: [[], [0], [3], [0, 3], [-5], [0, -5], [3, -5], [0, 3, -5]]

Three trims give eight presets, each keeping the bank order of the trims it engages.

Constraints

1 <= offsets.length <= 10
-10 <= offsets[i] <= 10
All values in offsets are different

The values you return may be in any order.

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 all_presets(offsets: list[int]) -> list[list[int]]:

Java

public List<List<Integer>> allPresets(int[] offsets)

The intended approach

Time O(n * 2^n)Space O(n) beyond the returned catalogue

Recurse over the bank one trim at a time, first exploring the branch that leaves the trim idle and then the branch that engages it, pushing and popping a shared partial preset. Because the recursion always advances left to right, whatever is on the partial list is already in bank order when a leaf is recorded.

Pythonentry point: all_presets

def all_presets(offsets):
    result = []
    chosen = []
    n = len(offsets)

    def walk(i):
        if i == n:
            result.append(list(chosen))
            return
        walk(i + 1)
        chosen.append(offsets[i])
        walk(i + 1)
        chosen.pop()

    walk(0)
    return result

The obvious version

Count from 0 to 2^n - 1 and read each counter value as an engage/idle pattern over the bank.

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_all_presets(offsets):
    n = len(offsets)
    out = []
    for mask in range(2 ** n):
        pick = []
        for i in range(n):
            if (mask >> i) % 2 == 1:
                pick.append(offsets[i])
        out.append(pick)
    return out

How this problem was checked

Reference solution executed against 7 test cases — 7 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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