Given an array of integers
nums, half of the integers in
nums are odd, and the other half are even.
Sort the array so that whenever
nums[i] is odd,
i is odd, and whenever
nums[i] is even,
i is even.
Return any answer array that satisfies this condition.
Input: nums = [4,2,5,7] Output: [4,5,2,7] Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
Input: nums = [2,3] Output: [2,3]
2 <= nums.length <= 2 * 104
- Half of the integers in
0 <= nums[i] <= 1000
Follow Up: Could you solve it in-place?
Iterate the list and store the odd, even numbers separately. And then created a new empty list and based on indices to add odd, even numbers.
class Solution: def sortArrayByParityII(self, nums: List[int]) -> List[int]: results =  odds =  evens =  for num in nums: if num % 2 == 0: evens.append(num) else: odds.append(num) i = 0 while i < len(nums): if i % 2 == 0: value = evens.pop(0) else: value = odds.pop(0) results.append(value) i += 1 return results
- Time Complexity: O(N).
- Space Complexity: O(N).