In this guide, you will get 922. Sort Array By Parity II LeetCode Solution with the best time and space complexity. The solution to Sort Array By Parity II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
Example 1:
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.
Example 2:
Input: nums = [2,3]
Output: [2,3]
Constraints:
2 <= nums.length <= 2 * 104
nums.length is even.
Half of the integers in nums are even.
0 <= nums[i] <= 1000
Follow Up: Could you solve it in-place?
Complexity Analysis
Time Complexity:
Space Complexity:
922. Sort Array By Parity II LeetCode Solution in C++
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
const int n = nums.size();
for (int i = 0, j = 1; i < n; i += 2, j += 2) {
while (i < n && nums[i] % 2 == 0)
i += 2;
while (j < n && nums[j] % 2 == 1)
j += 2;
if (i < n)
swap(nums[i], nums[j]);
}
return nums;
}
}; /* code provided by PROGIEZ */
922. Sort Array By Parity II LeetCode Solution in Java
class Solution {
public int[] sortArrayByParityII(int[] nums) {
final int n = nums.length;
for (int i = 0, j = 1; i < n; i += 2, j += 2) {
while (i < n && nums[i] % 2 == 0)
i += 2;
while (j < n && nums[j] % 2 == 1)
j += 2;
if (i < n) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
return nums;
}
} // code provided by PROGIEZ
922. Sort Array By Parity II LeetCode Solution in Python
class Solution:
def sortArrayByParityII(self, nums: list[int]) -> list[int]:
n = len(nums)
i = 0
j = 1
while i < n:
while i < n and nums[i] % 2 == 0:
i += 2
while j < n and nums[j] % 2 == 1:
j += 2
if i < n:
nums[i], nums[j] = nums[j], nums[i]
return nums # code by PROGIEZ
