<p>Given an array of integers <code>nums</code>, half of the integers in <code>nums</code> are <strong>odd</strong>, and the other half are <strong>even</strong>.</p>

<p>Sort the array so that whenever <code>nums[i]</code> is odd, <code>i</code> is <strong>odd</strong>, and whenever <code>nums[i]</code> is even, <code>i</code> is <strong>even</strong>.</p>

<p>Return <em>any answer array that satisfies this condition</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,2,5,7]
<strong>Output:</strong> [4,5,2,7]
<strong>Explanation:</strong> [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3]
<strong>Output:</strong> [2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>nums.length</code> is even.</li>
	<li>Half of the integers in <code>nums</code> are even.</li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow Up:</strong> Could you solve it in-place?</p>

class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& nums) {
       int n=nums.size();
       
       int i=0,j=1;
       while(i<n && j<n){
        if(nums[i]%2==0 && nums[j]%2!=0){
            i+=2;
            j+=2;
        }
        else if(nums[i]%2!=0 && nums[j]%2==0){
            swap(nums[i],nums[j]);
        }
        else if(nums[i]%2==0){
            i+=2;
        }
        else if(nums[j]%2!=0){
            j+=2;
        }
       }
       return nums;
    }
};
23800