0260EasyMathBit ManipulationRecursion

Tally in the Doubling 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 press opens a run with one sheet in the stack and doubles the stack on every pass after that, so the stack sizes the run produces are 1, 2, 4, 8, 16 and onwards without limit.

You are handed tally, a single line lifted from the shop ledger. A ledger line may be zero, and it may be negative, because a reversal is written as a negative count.

Return true when some pass of the doubling run leaves exactly tally sheets in the stack, and false otherwise.

Examples

Example 1

Input: tally = 64
Output: true

Doubling from a single sheet gives stacks of 1, 2, 4, 8, 16, 32 and then 64, so the seventh pass of the run leaves exactly this many sheets.

Example 2

Input: tally = 6
Output: false

The run leaves 4 sheets on one pass and 8 on the next, so it never leaves 6.

Example 3

Input: tally = -64
Output: false

Every stack the run produces holds at least one sheet, so a negative ledger line is not one of them.

Constraints

-2^31 <= tally <= 2^31 - 1

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 in_doubling_run(tally: int) -> bool:

Java

public boolean inDoublingRun(int tally)

The intended approach

Time O(1): one comparison and one bitwise and.Space O(1).

A stack size the run produces is a positive number whose binary form carries a single 1 bit. Subtracting one from such a number clears that bit and sets every bit below it, so the two values share no bit at all and their bitwise and is zero. Zero and the negative lines are ruled out up front, since the run never produces them.

Pythonentry point: in_doubling_run

def in_doubling_run(tally):
    return tally > 0 and tally & (tally - 1) == 0

The obvious version

Replay the run: start at one sheet and keep doubling until the stack reaches or passes the tally, then check whether it landed on it. O(log tally).

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_in_doubling_run(tally):
    if tally <= 0:
        return False
    stack = 1
    while stack < tally:
        stack *= 2
    return stack == tally

How this problem was checked

Reference solution executed against 22 test cases — 22 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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