1076EasyArrayTwo PointersSorting

Evens to the Front

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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 list of whole numbers reads values.

Return the list rearranged so that every even value comes before every odd one, with the even values keeping the order they had among themselves and the odd values likewise keeping theirs.

Examples

Example 1

Input: values = [1, 2, 3, 4]
Output: [2, 4, 1, 3]

The evens are 2 and 4 in that order, and the odds are 1 and 3 in that order, so the evens go first with each group's order untouched.

Example 2

Input: values = [7, 7, 2]
Output: [2, 7, 7]

The only even value moves to the front and the two odd sevens follow in the order they were in.

Example 3

Input: values = [5]
Output: [5]

A single odd value has nothing to move ahead of it.

Constraints

1 <= values.length <= 5000
0 <= values[i] <= 5000

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 sort_array_by_parity(values: list[int]) -> list[int]:

Java

public int[] sortArrayByParity(int[] values)

The intended approach

Time O(n) over the list's length.Space O(n) for the answer.

Since each group keeps its own order, the answer is simply the evens written out in the order they appear, followed by the odds in the order they appear. So walk the list twice, or once collecting into two lists and then joining them. Either way each value is looked at a fixed number of times, and nothing is compared with anything else, which is what makes this cheaper than sorting. Whether a value is even is settled by its remainder on halving.

Pythonentry point: sort_array_by_parity

def sort_array_by_parity(values):
    evens = []
    odds = []
    for value in values:
        if value % 2 == 0:
            evens.append(value)
        else:
            odds.append(value)
    return evens + odds

The obvious version

Sort the positions instead, ordering them by whether the value there is odd and then by the position itself, which keeps each group's own order while pushing the odds to the back. Then read the values off in that order. Costs a sort rather than two walks.

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_sort_array_by_parity(values):
    spots = list(range(len(values)))
    spots.sort(key=lambda at: (values[at] % 2, at))
    out = []
    for at in spots:
        out.append(values[at])
    return out

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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