Array With Elements Not Equal to Average of Neighbors
<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of <strong>distinct</strong> integers. You want to rearrange the elements in the array such that every element in the rearranged array is <strong>not</strong> equal to the <strong>average</strong> of its neighbors.</p>

<p>More formally, the rearranged array should have the property such that for every <code>i</code> in the range <code>1 &lt;= i &lt; nums.length - 1</code>, <code>(nums[i-1] + nums[i+1]) / 2</code> is <strong>not</strong> equal to <code>nums[i]</code>.</p>

<p>Return <em><strong>any</strong> rearrangement of </em><code>nums</code><em> that meets the requirements</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> [1,2,4,5,3]
<strong>Explanation:</strong>
When i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5.
When i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5.
When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [6,2,0,9,7]
<strong>Output:</strong> [9,7,6,2,0]
<strong>Explanation:</strong>
When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.
Note that the original array [6,2,0,9,7] also satisfies the conditions.</pre>

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

<ul>
	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Greedy; Array; Sorting
class Solution {
public:
    vector<int> rearrangeArray(vector<int>& nums) {
        ranges::sort(nums);
        vector<int> ans;
        int n = nums.size();
        int m = (n + 1) >> 1;
        for (int i = 0; i < m; ++i) {
            ans.push_back(nums[i]);
            if (i + m < n) {
                ans.push_back(nums[i + m]);
            }
        }
        return ans;
    }
};
