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easyBasicsSometimes asked

Separate Odd and Even

Basics · 6 min

Given an integer array nums, return a pair of lists: all even numbers and all odd numbers, each in their original relative order.

In plain English

One pass; route each element into the even or odd output list.

Examples

Example 1

Input: nums = [1, 2, 3, 4]
Output: [[2, 4], [1, 3]]
Note: Evens then odds.

Example 2

Input: nums = [2, 4]
Output: [[2, 4], []]
Note: All even.

Example 3

Input: nums = [1]
Output: [[], [1]]
Note: All odd.

Constraints

0 ≤ nums.length ≤ 10^5

Clarifying questions2First question: Must relative order be preserved?Explore questions

Q1.Must relative order be preserved?
Yes — appending in scan order keeps it stable.
Q2.In place or two lists?
Two lists for clarity; in-place is possible if unstable order is acceptable.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Create two lists; append each element to the even or odd list based on `x % 2 == 0`. Order is preserved by appending as you scan.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you produce a stable partition and understand the space/order trade-offs.

Where candidates slip

Reversing a group, or the negative-parity bug.

How to stand out

Contrast stable two-list output with the unstable in-place two-pointer swap.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Route into two lists

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Two output buckets — what decides which one an element goes to?

2Hint 2Explore hint

Appending as you scan preserves each group's order automatically.

3Hint 3Explore hint

Use `% 2 == 0` for the even bucket.

A · Core idea

Route each element into an even or odd output list based on its parity. Appending in scan order makes the partition stable — each group keeps its original relative order — for free.

Why appending preserves order. You read left to right and append to the back of the matching list, so within each bucket the elements appear in the same sequence they had in the input. No sorting or extra bookkeeping is needed.

Two output lists vs in place. Producing two separate lists is O(n) extra space but simplest. An in-place stable partition is harder; an unstable in-place partition (two-pointer swap) is O(1) space but reorders within groups — mention the trade-off.

Parity robustness. Test % 2 == 0 so negative evens are bucketed correctly.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

nums = [1, 2, 3, 4]

odd

Stateeven=[], odd=[1]

Final result[[2, 4], [1, 3]]

C · Code with teaching comments

solution.javaJava

Route into two lists — optimal

1
public java.util.List<java.util.List<Integer>> separate(int[] nums) {
2
    java.util.List<Integer> even = new java.util.ArrayList<>();
3
    java.util.List<Integer> odd = new java.util.ArrayList<>();
4
    for (int x : nums) {
5
        if (x % 2 == 0) even.add(x);
6
        else odd.add(x);
7
    }
8
    return java.util.List.of(even, odd);
9
}

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(n) space — one parity test and one append per element; two lists hold n elements total, so O(n) space.

E · Edge cases and pitfalls5nums = [2, 4] — All even → [[2, 4], []].Explore edge cases

Edge cases handled

nums = [2, 4]All even → [[2, 4], []].
nums = [1]All odd → [[], [1]].
nums = []→ [[], []].

Pitfalls

Inserting at the front and reversing a group's order.
Parity bug on negative numbers.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic java.util.List<java.util.List<Integer>> separate(int[] nums) { java.util.List<Integer> even = new java.util.ArrayList<>(); java.util.List<Integer> odd = new java.util.ArrayList<>(); for (int x : nums) { if (x % 2 == 0) even.add(x);Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Reversing one group by inserting at the front.
Parity bug on negative numbers.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Stable partition into buckets

Code pattern to memorize

even=[x for even]; odd=[x for odd]   (stable by append)

Rules to remember

1Append in scan order.
2Parity picks the bucket.
3Order preserved per group.

Use this pattern when

Splitting into groups while preserving order

Interview takeaway

Appending while you scan preserves relative order automatically.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Stable partition into buckets. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Sort Array By Parity
Coming soon
Evens before odds in one array (order can be relaxed).
Move Zeros to End
A related stable partition by a predicate.

easyBasicsSometimes asked

Separate Odd and Even

Basics · 6 min

Given an integer array nums, return a pair of lists: all even numbers and all odd numbers, each in their original relative order.

In plain English

One pass; route each element into the even or odd output list.

Examples

Example 1

Input: nums = [1, 2, 3, 4]
Output: [[2, 4], [1, 3]]
Note: Evens then odds.

Example 2

Input: nums = [2, 4]
Output: [[2, 4], []]
Note: All even.

Example 3

Input: nums = [1]
Output: [[], [1]]
Note: All odd.

Constraints

0 ≤ nums.length ≤ 10^5

Clarifying questions2First question: Must relative order be preserved?Explore questions

Q1.Must relative order be preserved?
Yes — appending in scan order keeps it stable.
Q2.In place or two lists?
Two lists for clarity; in-place is possible if unstable order is acceptable.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Create two lists; append each element to the even or odd list based on `x % 2 == 0`. Order is preserved by appending as you scan.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you produce a stable partition and understand the space/order trade-offs.

Where candidates slip

Reversing a group, or the negative-parity bug.

How to stand out

Contrast stable two-list output with the unstable in-place two-pointer swap.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Route into two lists

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Two output buckets — what decides which one an element goes to?

2Hint 2Explore hint

Appending as you scan preserves each group's order automatically.

3Hint 3Explore hint

Use `% 2 == 0` for the even bucket.

A · Core idea

Route each element into an even or odd output list based on its parity. Appending in scan order makes the partition stable — each group keeps its original relative order — for free.

Parity robustness. Test % 2 == 0 so negative evens are bucketed correctly.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

nums = [1, 2, 3, 4]

odd

Stateeven=[], odd=[1]

Final result[[2, 4], [1, 3]]

C · Code with teaching comments

solution.javaJava

Route into two lists — optimal

1
public java.util.List<java.util.List<Integer>> separate(int[] nums) {
2
    java.util.List<Integer> even = new java.util.ArrayList<>();
3
    java.util.List<Integer> odd = new java.util.ArrayList<>();
4
    for (int x : nums) {
5
        if (x % 2 == 0) even.add(x);
6
        else odd.add(x);
7
    }
8
    return java.util.List.of(even, odd);
9
}

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(n) space — one parity test and one append per element; two lists hold n elements total, so O(n) space.

E · Edge cases and pitfalls5nums = [2, 4] — All even → [[2, 4], []].Explore edge cases

Edge cases handled

nums = [2, 4]All even → [[2, 4], []].
nums = [1]All odd → [[], [1]].
nums = []→ [[], []].

Pitfalls

Inserting at the front and reversing a group's order.
Parity bug on negative numbers.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Reversing one group by inserting at the front.
Parity bug on negative numbers.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Stable partition into buckets

Code pattern to memorize

even=[x for even]; odd=[x for odd]   (stable by append)

Rules to remember

1Append in scan order.
2Parity picks the bucket.
3Order preserved per group.

Use this pattern when

Splitting into groups while preserving order

Interview takeaway

Appending while you scan preserves relative order automatically.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Stable partition into buckets. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Sort Array By Parity
Coming soon
Evens before odds in one array (order can be relaxed).
Move Zeros to End
A related stable partition by a predicate.