922. Sort Array By Parity II(Solution || Leetcode easy || Java)

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?

SOLUTION(1):

class Solution {
public int[] sortArrayByParityII(int[] nums) {
//first we will seperate even and odd numbers and store them in a temporary array
int i = 0;
int j = nums.length — 1;

int[] temp = new int[nums.length];

for(int k = 0; k < nums.length; k++){
if(nums[k]%2 == 0){
temp[i] = nums[k];
i++;
}
else{
temp[j] = nums[k];
j — ;
}
}

//now we are arranging even num at even index and odd at odd index
int[] ans = new int[nums.length];

int even = 0;
int odd = j + 1;

for(int k = 0; k < nums.length; k++){
if(k%2 == 0){
ans[k] = temp[even];
even++;
}
else{
ans[k] = temp[odd];
odd++;
}
}

return ans;
}
}

SOLUTION(2):

class Solution {
public int[] sortArrayByParityII(int[] nums) {
//Initilaise varibale for even index
int i = 0;

//Initialise variable for odd index
int j = 1;

//Intialise array to return
int[] ans = new int[nums.length];

for(int k = 0; k < nums.length; k++){
//if element is even, add it in ans array at the even index
if(nums[k]%2 == 0){
ans[i] = nums[k];
i += 2;
}
//if element is odd, add that element in the ans array at odd index
else{
ans[j] = nums[k];
j += 2;
}
}
return ans;
}
}

Runtime: 2 ms, faster than 99.88% of Java online submissions for Sort Array By Parity II.

Memory Usage: 43.9 MB, less than 96.22% of Java online submissions for Sort Array By Parity II.

Thank you for reading. If you have any queries then, plese let me know in the comment section. I will surely be responsive toward it.