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

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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.