How do you solve Transform Array by Parity on LeetCode?

Transform Array by Parity (LeetCode #3467) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transforming numbers can be done with counting instead of sorting.

What is the time complexity of Transform Array by Parity?

The optimal solution for Transform Array by Parity runs in O(n) time and uses O(n) space. Counting evens and odds takes O(n), and constructing the result also takes O(n), leading to O(n) overall.

What is the brute force solution for Transform Array by Parity?

The brute force approach for Transform Array by Parity is Brute Force, with O(n²) time complexity and O(1) space complexity. We can transform the array by iterating through it, replacing even numbers with 0 and odd numbers with 1. Then, we sort the resulting array. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Transform Array by Parity?

Transform Array by Parity uses the following concepts: Array, Sorting, Counting. The recommended patterns to study are: Counting, Array.

What are the interview tips for Transform Array by Parity?

Explain your thought process clearly. Consider edge cases like all evens or all odds. Optimize before coding.

What are common mistakes when solving Transform Array by Parity?

Forgetting to count evens and odds correctly. Confusing the order of operations.

#3467

Transform Array by Parity

Easy

ArraySortingCountingCountingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Count the even and odd numbers, then construct the result directly based on counts, which avoids sorting.

⚙️

Algorithm

3 steps

1Step 1: Initialize counters for even and odd numbers.
2Step 2: Iterate through the array to count evens and odds.
3Step 3: Construct the result array with '0's followed by '1's based on the counts.

solution.py3 lines

1def transform_array(nums):
2    evens = sum(1 for x in nums if x % 2 == 0)
3    return [0] * evens + [1] * (len(nums) - evens)

ℹ

Complexity note: Counting evens and odds takes O(n), and constructing the result also takes O(n), leading to O(n) overall.

1Transforming numbers can be done with counting instead of sorting.
2Directly constructing the output array saves time.

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