How do you solve Sort Array By Parity on LeetCode?

Sort Array By Parity (LeetCode #905) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a two-pointer technique can significantly reduce the time complexity.

What is the brute force solution for Sort Array By Parity?

The brute force approach for Sort Array By Parity is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating two separate lists: one for even numbers and one for odd numbers. We then concatenate these lists to achieve the desired order. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sort Array By Parity?

Sort Array By Parity uses the following concepts: Array, Two Pointers, Sorting. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Sort Array By Parity?

Always clarify if the order of even and odd numbers matters. Discuss your thought process and reasoning for choosing an approach. Be prepared to explain the time and space complexities of your solution.

What are common mistakes when solving Sort Array By Parity?

Not considering edge cases like arrays with all even or all odd numbers. Failing to swap correctly and ending up with incorrect placements.

#905

Sort Array By Parity

Easy

ArrayTwo PointersSortingTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to rearrange the numbers in a single pass, which is more efficient. One pointer will track the position to place even numbers, while the other will iterate through the array.

⚙️

Algorithm

5 steps

1Step 1: Initialize a pointer 'evenIndex' at the start of the array.
2Step 2: Iterate through the array with a second pointer 'i'.
3Step 3: If the current number is even, swap it with the number at 'evenIndex' and increment 'evenIndex'.
4Step 4: Continue until the end of the array.
5Step 5: Return the modified array.

solution.py10 lines

1# Full working Python code
2
3def sortArrayByParity(nums):
4    evenIndex = 0
5    for i in range(len(nums)):
6        if nums[i] % 2 == 0:
7            nums[i], nums[evenIndex] = nums[evenIndex], nums[i]
8            evenIndex += 1
9    return nums
10

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(1) since we are rearranging the elements in place without using additional space.

1Using a two-pointer technique can significantly reduce the time complexity.
2In-place modifications help save space and improve efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.