0015EasyArrayTwo Pointers

Compact Blank Cells

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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.

One column of a spreadsheet is held in the integer array cells. A cell nobody filled in is stored as 0; any other entry is a signed amount somebody typed, which may be negative.

Pack the typed amounts against the front of the column and push every blank behind them. The typed amounts have to come out in the same succession they went in, so only the blanks change position relative to anything else.

Rearrange cells itself using no more than a constant amount of extra space, then return cells.

Examples

Example 1

Input: cells = [4, 0, 5, 0, -3]
Output: [4, 5, -3, 0, 0]

The typed amounts still run 4, then 5, then -3, and the two blanks gather behind them.

Example 2

Input: cells = [0, -6, 0, 0, 3, 0, -1]
Output: [-6, 3, -1, 0, 0, 0, 0]

Four blanks slide back while -6, 3 and -1 hold the order they started in.

Example 3

Input: cells = [9, -2, 7]
Output: [9, -2, 7]

Nothing is blank here, so the column already satisfies the layout and comes back unchanged.

Constraints

1 <= cells.length <= 10^4
-2^31 <= cells[i] <= 2^31 - 1
Only a constant amount of extra space may be used

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 compact_blanks(cells: list[int]) -> list[int]:

Java

public int[] compactBlanks(int[] cells)

The intended approach

Time O(n)Space O(1)

Keep a write cursor for the next front slot. Read the column left to right and copy each typed amount to the cursor, which never overtakes the reader, so the succession survives. Whatever sits from the cursor to the end after that pass is surplus, so overwrite it with blanks.

Pythonentry point: compact_blanks

def compact_blanks(cells):
    write = 0
    for value in cells:
        if value != 0:
            cells[write] = value
            write += 1
    for i in range(write, len(cells)):
        cells[i] = 0
    return cells

The obvious version

Work on a copy and keep sweeping it, swapping any blank that sits directly in front of a typed amount, until a full sweep changes nothing. Swaps never reorder two typed amounts. O(n^2).

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_compact_blanks(cells):
    data = list(cells)
    shifted = True
    while shifted:
        shifted = False
        for i in range(len(data) - 1):
            if data[i] == 0 and data[i + 1] != 0:
                data[i], data[i + 1] = data[i + 1], data[i]
                shifted = True
    return data

How this problem was checked

Reference solution executed against 9 test cases — 9 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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