1079EasyArrayTwo Pointers

Doubling the Blanks on a Fixed Strip

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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 strip of digits reads strip. Every 0 on it is a blank.

Double each blank in place: each 0 becomes two 0s and everything after it shifts one place along. The strip keeps its original length, so whatever is pushed off the end is lost.

Return the strip afterwards.

Examples

Example 1

Input: strip = [0, 1, 2]
Output: [0, 0, 1]

The single blank becomes two, pushing everything along, and the last entry falls off the end.

Example 2

Input: strip = [5, 6, 7]
Output: [5, 6, 7]

There is no blank anywhere, so nothing moves.

Example 3

Input: strip = [0, 0]
Output: [0, 0]

The first blank doubles into both places, and the second blank's copies both fall off the end.

Constraints

1 <= strip.length <= 10^4
0 <= strip[i] <= 9

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 duplicate_zeros(strip: list[int]) -> list[int]:

Java

public int[] duplicateZeros(int[] strip)

The intended approach

Time O(n) over the strip's length.Space O(1) beyond the strip itself.

Writing forwards would overwrite entries not yet read, so write backwards instead. First count the blanks, which tells how much longer the strip would be if nothing were lost: the original length plus that count. Then walk backwards from the end of that imagined longer strip, taking entries off the back of the original. Each entry is written at the imagined position when that position still falls inside the real strip, and a blank is written twice, once for each of the two positions it now claims. Because the imagined position is never behind the position being read, nothing is overwritten before it is used. One pass to count and one to write.

Pythonentry point: duplicate_zeros

def duplicate_zeros(strip):
    out = list(strip)
    n = len(out)
    blanks = 0
    for value in out:
        if value == 0:
            blanks += 1
    write = n + blanks - 1
    for read in range(n - 1, -1, -1):
        if write < n:
            out[write] = out[read]
        write -= 1
        if out[read] == 0:
            if write < n:
                out[write] = 0
            write -= 1
    return out

The obvious version

Build the answer somewhere else. Walk the strip forwards appending each entry to a fresh list, appending a second copy whenever the entry is a blank, and stop once the fresh list has reached the original length. Spends a second strip's worth of room rather than writing backwards in place.

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_duplicate_zeros(strip):
    n = len(strip)
    built = []
    for value in strip:
        built.append(value)
        if len(built) == n:
            break
        if value == 0:
            built.append(0)
            if len(built) == n:
                break
    return built

How this problem was checked

Reference solution executed against 15 test cases — 15 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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