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Tyler Scott Williams
Tyler Scott Williams

Posted on • Updated on • Originally published at

Sort Array by Parity

This post is part of my "LeetCode for 1x Developers" series, in which I struggle through LeetCode problems. Sometimes I figure it out, other times I don't. Either way, I give each problem my best shot, and write up my thought process through the challenges

Problem description

Problem on leetcode

Given an array A of non-negative integers, half of the integers in A ar odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd. And when A[i] is even, i is even.

You may return any answer array that satisfies this conditions.


Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,6], [2,7,4,5] would also have been accepted.



OK, so it looks like we don't have to actually sort the array in any meaningful way. We can probably do something with it in-place. We also know we'll always have enough numbers to fill the pattern, since half of the integers are odd and half are even.

Can we do this in place? Let's look at the example:

[4, 2, 5, 7]

Check A[0]. It's 4. OK, it's all good since i and A[i] are both even.

Now check A[2], it's 2. i is odd, but A[i] is even. We need to move it. I almost want to just swap it with the next value, but I am pretty sure that won't work all the time so I'm going to leave it alone.

What if we stored all of the even numbers in an array called even and all the odd numbers in an array called odd? We could then rebuild an array and pull a number from each depending on the parity of the number.

I'm worried about basically running two loops and using three arrays. But I think it's still technically O(n), and the space constraints will only ever be 2n. Maybe that's reasonable. Let's find out.

Here's my first pass at expressing this solution in JavaScript (ES6):

var sortArrayByParityII = function(A) {
    let even = [];
    let odd = [];
    let result = [];
    for (let i=0; i<A.length; i++) {
        if (A[i] % 2 === 0) {
        } else {
    for (let j=0; j<A.length; j++) {
        if (j % 2 === 0) {
            result[j] = even.pop();
        } else {
            result[j] = odd.pop();
    return result;
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It passed, faster than 51.97% of JS solutions, and with less than 27.27% of JS solutions. I wonder where I can optimize it. It feels like I should be able to do something in one loop, but I can't really reckon with it.


After checking the discussion section, I found there is a one-loop solution. It's not too far off from my first pass.

First, you set up an empty array, which you'll use as the result. Call it result.

Then you set an evenIndex to 0, and an oddIndex to `.

You loop through the input array A. If A[i] is even, you set result[evenIndex] to the value and increment evenIndex by two. If it's odd, you set result[oddINdex] to the value and increment oddIndex by two.

You can express it like this:

var sortArrayByParityII = function(A) {
let result = [];
let evenIndex = 0;
let oddIndex = 1;
for (let i=0; i<A.length; i++) {
if (A[i] % 2 === 0) {
result[evenIndex] = A[i];
evenIndex += 2;
} else {
result[oddIndex] = A[i];
oddIndex += 2;
return result;

And it runs faster than 86% of JS solutions with less memory than 36% of other solutions. Nice!

After having done a dozen or so easy problems, I'm really starting to see most Leetcode easy problems as array mapping solutions.

Top comments (1)

ilamparithi10 profile image

Good one :)