<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) {
        vector<int> odd;
        vector<int> even;
        for(int n:nums){
            if(n%2 ==0){
                even.push_back(n);
            }else{
                odd.push_back(n);
            }
        }
        sort(odd.begin(),odd.end());
        sort(even.begin(),even.end());
        int e=0;
        int o=0;
        for(int i=0;i<nums.size();i++){
            if(i%2==0){
                nums[i] = even[e];
                e++;
            }else{
                nums[i] = odd[o];
                o++;
            }
        }
        return nums;
    }
};
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