[Arrays] Sort Array By Parity

📋 This is my note-taking from what I learned in LeetCode!

Problem

Problem Link

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.

Return any array that satisfies this condition.

Example 1:

• Input: nums = [3,1,2,4]
• Output: [2,4,3,1]
• Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Example 2:

• Input: nums = [0]
• Output: [0]

Constraints:

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

Solution

Solution Link

Approach 1: Sort

Use a custom comparator when sorting, to sort by parity.

Algorithm:

1. Sort the input list “nums” based on the remainder of each element when divided by 2. The key function used for sorting is a lambda function that returns the result of x%2 for each element x in the list.
2. Return the sorted list as the output. The sorted list will have all the even numbers first, followed by all the odd numbers.

def sortArrayByParity(nums: list[int]) -> list[int]:
    nums.sort(key=lambda x: x % 2)
    return nums

nums = [3, 1, 2, 4]
sorted_nums = sortArrayByParity(nums)
print(sorted_nums)

Complexity Analysis

• Time Complexity: O(Nlog⁡N), where N is the length of A.
• Space Complexity: O(N) for the sort, depending on the built-in implementation of sort.

Approach 2: Two Pass

Algorithm:

def sortArrayByParity(nums: list[int]) -> list[int]:
    return [x for x in nums if x % 2 == 0] + [x for x in nums if x % 2 == 1]

nums = [3, 1, 2, 4]
sorted_nums = sortArrayByParity(nums)
print(sorted_nums)

Complexity Analysis

• Time Complexity: O(N), where N is the length of A.
• Space Complexity: O(N), the space used by the answer.

Approach 3: In-Place

If we want to do the sort in-place, we can use quicksort, a standard textbook algorithm.

Algorithm:

We’ll maintain two pointers i and j. The loop invariant is everything below i has parity 0 (ie. A[k] % 2 == 0 when k < i), and everything above j has parity 1.

Then, there are 4 cases for (A[i] % 2, A[j] % 2):

• If it is (0, 1), then everything is correct: i++ and j–.
• If it is (1, 0), we swap them so they are correct, then continue.
• If it is (0, 0), only the i place is correct, so we i++ and continue.
• If it is (1, 1), only the j place is correct, so we j– and continue.

Throughout all 4 cases, the loop invariant is maintained, and j-i is getting smaller. So eventually we will be done with the array sorted as desired.

def sortArrayByParity(nums: list[int]) -> list[int]:
    i, j = 0, len(nums) - 1
    while i < j:
        if nums[i] % 2 > nums[j] % 2:
            nums[i], nums[j] = nums[j], nums[i]
        if nums[i] % 2 == 0:
            i += 1
        if nums[j] % 2 == 1:
            j -= 1
    return nums

nums = [3, 1, 2, 4]
sorted_nums = sortArrayByParity(nums)
print(sorted_nums)

Complexity Analysis

• Time Complexity: O(N), where N is the length of A. Each step of the while loop makes j-i decrease by at least one. (Note that while quicksort is O(Nlog⁡N) normally, this is O(N) because we only need one pass to sort the elements.)
• Space Complexity: O(1) in additional space complexity.

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