Sort Even and Odd Indices Independently
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. Rearrange the values of <code>nums</code> according to the following rules:</p>

<ol>
	<li>Sort the values at <strong>odd indices</strong> of <code>nums</code> in <strong>non-increasing</strong> order.

    <ul>
    	<li>For example, if <code>nums = [4,<strong><u>1</u></strong>,2,<u><strong>3</strong></u>]</code> before this step, it becomes <code>[4,<u><strong>3</strong></u>,2,<strong><u>1</u></strong>]</code> after. The values at odd indices <code>1</code> and <code>3</code> are sorted in non-increasing order.</li>
    </ul>
    </li>
    <li>Sort the values at <strong>even indices</strong> of <code>nums</code> in <strong>non-decreasing</strong> order.
    <ul>
    	<li>For example, if <code>nums = [<u><strong>4</strong></u>,1,<u><strong>2</strong></u>,3]</code> before this step, it becomes <code>[<u><strong>2</strong></u>,1,<u><strong>4</strong></u>,3]</code> after. The values at even indices <code>0</code> and <code>2</code> are sorted in non-decreasing order.</li>
    </ul>
    </li>

</ol>

<p>Return <em>the array formed after rearranging the values of</em> <code>nums</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [4,1,2,3]
<strong>Output:</strong> [2,3,4,1]
<strong>Explanation:</strong> 
First, we sort the values present at odd indices (1 and 3) in non-increasing order.
So, nums changes from [4,<strong><u>1</u></strong>,2,<strong><u>3</u></strong>] to [4,<u><strong>3</strong></u>,2,<strong><u>1</u></strong>].
Next, we sort the values present at even indices (0 and 2) in non-decreasing order.
So, nums changes from [<u><strong>4</strong></u>,1,<strong><u>2</u></strong>,3] to [<u><strong>2</strong></u>,3,<u><strong>4</strong></u>,1].
Thus, the array formed after rearranging the values is [2,3,4,1].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,1]
<strong>Output:</strong> [2,1]
<strong>Explanation:</strong> 
Since there is exactly one odd index and one even index, no rearrangement of values takes place.
The resultant array formed is [2,1], which is the same as the initial array. 
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>
Array; Sorting
class Solution {
public:
    vector<int> sortEvenOdd(vector<int>& nums) {
        int n = nums.size();
        vector<int> a;
        vector<int> b;
        for (int i = 0; i < n; ++i) {
            if (i % 2 == 0) {
                a.push_back(nums[i]);
            } else {
                b.push_back(nums[i]);
            }
        }
        sort(a.begin(), a.end());
        sort(b.rbegin(), b.rend());
        vector<int> ans(n);
        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) {
            ans[i] = a[j];
        }
        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) {
            ans[i] = b[j];
        }
        return ans;
    }
};