0182MediumStringDynamic Programming

Split The Swatch Tape

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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 colour lab keeps twenty-six swatches, numbered 1 through 26. When a batch is mixed, the machine punches the swatch numbers onto a paper tape one after another with no separator between them, so a batch of swatch 8 then swatch 15 comes out as the tape "815".

You are handed a finished tape as the digit string tape and have to work out how many batches could have punched it. A swatch number is punched with no padding, so no number on the tape may begin with 0, and every number must land in the range 1 to 26.

Return the count of batches that produce tape. Return 0 when the tape cannot have come from any batch.

Examples

Example 1

Input: tape = "2613"
Output: 4

The tape splits as 2,6,1,3 or 26,1,3 or 2,6,13 or 26,13, so four batches fit.

Example 2

Input: tape = "907"
Output: 0

The middle 0 cannot stand alone and 90 is past 26, so nothing accounts for it.

Example 3

Input: tape = "1120"
Output: 2

The trailing 0 has to join the 2 in front of it as 20, leaving 1,1,20 or 11,20.

Constraints

1 <= tape.length <= 100
tape contains only digits and may begin with one or more zeros.

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 count_tape_readings(tape: str) -> int:

Java

public int countTapeReadings(String tape)

The intended approach

Time O(n)Space O(1)

The batches ending at position i split into those whose last swatch number took one digit and those whose last number took two, so the count at i is the count at i-1 plus the count at i-2, each guarded. A single digit only counts when it is not 0, and a pair only counts when it starts with 1, or starts with 2 and ends at 6 or below. Only the last two counts are ever needed, so keep two running values.

Pythonentry point: count_tape_readings

def count_tape_readings(tape):
    # Rolling pair of counts instead of a full table.
    # two_back = readings of tape[:i-1], one_back = readings of tape[:i].
    # Time O(n), space O(1).
    two_back = 1
    one_back = 0 if tape[0] == "0" else 1
    for i in range(1, len(tape)):
        current = 0
        if tape[i] != "0":
            current += one_back
        if tape[i - 1] == "1" or (tape[i - 1] == "2" and tape[i] <= "6"):
            current += two_back
        two_back, one_back = one_back, current
    return one_back

The obvious version

Recursively peel a swatch number off the front of the tape, trying one digit and then two digits whenever they are legal, and count the recursions that land exactly on the end. Explores every split, so it is exponential.

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_count_tape_readings(tape):
    def walk(i):
        if i == len(tape):
            return 1
        if tape[i] == "0":
            return 0
        total = walk(i + 1)
        if i + 1 < len(tape) and int(tape[i:i + 2]) <= 26:
            total += walk(i + 2)
        return total

    return walk(0)

How this problem was checked

Reference solution executed against 16 test cases — 16 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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