<p>Given an integer array <code>nums</code> of size <code>n</code>, return <em>the minimum number of moves required to make all array elements equal</em>.</p>

<p>In one move, you can increment or decrement an element of the array by <code>1</code>.</p>

<p>Test cases are designed so that the answer will fit in a <strong>32-bit</strong> integer.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
Only two moves are needed (remember each move increments or decrements one element):
[<u>1</u>,2,3]  =&gt;  [2,2,<u>3</u>]  =&gt;  [2,2,2]
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,10,2,9]
<strong>Output:</strong> 16
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
 public:
  int minMoves2(vector<int>& nums) {
    const int n = nums.size();
    nth_element(nums.begin(), nums.begin() + n / 2, nums.end());
    const int median = nums[n / 2];
    return accumulate(nums.begin(), nums.end(), 0, [&](int subtotal, int num) {
      return subtotal + abs(num - median);
    });
  }
};
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