To minimize the length of the array, we should maximize the number of operations performed.

To perform <code>k</code> operations, it is optimal to use the smallest <code>k</code> values and the largest <code>k</code> values in <code>nums</code>.

What is the best way to make pairs from the smallest <code>k</code> values and the largest <code>k</code> values so it is possible to remove all the pairs?

If we consider the smallest <code>k</code> values and the largest <code>k</code> values as two separate <strong>sorted 0-indexed</strong> arrays, <code>a</code> and <code>b</code>, It is optimal to pair <code>a[i]</code> and <code>b[i]</code>. So, a <code>k</code> is valid if <code>a[i] < b[i]</code> for all <code>i</code> in the range <code>[0, k - 1]</code>.

The greatest possible valid <code>k</code> can be found using binary search.

The answer is <code>nums.length - 2 * k</code>.