0451MediumStringBit ManipulationSimulation

Taps To Settle A Cell Register

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Trains the technique from

LeetCode 1404Number of Steps to Reduce a Number in Binary Representation to One

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 particle counter holds its running total in a row of cells given as the string cells, each cell either '0' or '1', highest place value first. The leading cell is always '1', so the row carries no padding, and the total may be far larger than a machine word.

An operator brings the total down to 1 with taps of a single key. Each tap does one of two things, and the operator has no choice about which:

when the total is even, the tap shifts it down to half its value;
when the total is odd, the tap raises it by one.

Return how many taps it takes to reach a total of 1. A row that already reads 1 needs none.

Examples

Example 1

Input: cells = "10011"
Output: 8

The row stands for 19. The taps run 19, 20, 10, 5, 6, 3, 4, 2, 1, which is eight taps in all.

Example 2

Input: cells = "1000"
Output: 3

The row stands for 8, and three taps bring it through 4 and 2 to 1.

Example 3

Input: cells = "1111"
Output: 5

The row stands for 15, and the taps run 15, 16, 8, 4, 2, 1.

Constraints

1 <= cells.length <= 500
cells consists only of the characters '0' and '1'
cells[0] == '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 taps_to_settle(cells: str) -> int:

Java

public int tapsToSettle(String cells)

The intended approach

Time O(n) for a row of n cells.Space O(1), just the tap count and the carry.

Sweep the cells from the low end towards the leading cell, holding a single carry that records whether the cells already passed have pushed a one into the current place. A place that reads one once the carry is folded in costs two taps and leaves a carry behind, while any other place costs one. If a carry is still live when the leading cell is reached, it adds one final tap.

Pythonentry point: taps_to_settle

def taps_to_settle(cells):
    # Read the register from its low end, carrying the effect of every nudge
    # leftwards, so the whole count comes out of one sweep.
    # O(n) time, O(1) extra space.
    taps = 0
    carry = 0
    for i in range(len(cells) - 1, 0, -1):
        bit = (1 if cells[i] == '1' else 0) + carry
        if bit == 1:
            # odd here: one nudge, and the shift that the nudge makes possible
            taps += 2
            carry = 1
        else:
            taps += 1
            carry = 1 if bit == 2 else 0
    if carry:
        taps += 1
    return taps

The obvious version

Turn the row into an arbitrary-precision integer and apply the two rules literally until the value is 1, counting the taps. O(n^2) digit work.

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_taps_to_settle(cells):
    value = 0
    for ch in cells:
        value = value * 2 + (1 if ch == '1' else 0)
    taps = 0
    while value > 1:
        if value % 2 == 0:
            value //= 2
        else:
            value += 1
        taps += 1
    return taps

How this problem was checked

Reference solution executed against 14 test cases — 14 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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