Minimum Adjacent Swaps to Make a Valid Array
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>

<p><strong>Swaps</strong> of <strong>adjacent</strong> elements are able to be performed on <code>nums</code>.</p>

<p>A <strong>valid</strong> array meets the following conditions:</p>

<ul>
	<li>The largest element (any of the largest elements if there are multiple) is at the rightmost position in the array.</li>
	<li>The smallest element (any of the smallest elements if there are multiple) is at the leftmost position in the array.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> swaps required to make </em><code>nums</code><em> a valid array</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,4,5,5,3,1]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Perform the following swaps:
- Swap 1: Swap the 3<sup>rd</sup> and 4<sup>th</sup> elements, nums is then [3,4,5,<u><strong>3</strong></u>,<u><strong>5</strong></u>,1].
- Swap 2: Swap the 4<sup>th</sup> and 5<sup>th</sup> elements, nums is then [3,4,5,3,<u><strong>1</strong></u>,<u><strong>5</strong></u>].
- Swap 3: Swap the 3<sup>rd</sup> and 4<sup>th</sup> elements, nums is then [3,4,5,<u><strong>1</strong></u>,<u><strong>3</strong></u>,5].
- Swap 4: Swap the 2<sup>nd</sup> and 3<sup>rd</sup> elements, nums is then [3,4,<u><strong>1</strong></u>,<u><strong>5</strong></u>,3,5].
- Swap 5: Swap the 1<sup>st</sup> and 2<sup>nd</sup> elements, nums is then [3,<u><strong>1</strong></u>,<u><strong>4</strong></u>,5,3,5].
- Swap 6: Swap the 0<sup>th</sup> and 1<sup>st</sup> elements, nums is then [<u><strong>1</strong></u>,<u><strong>3</strong></u>,4,5,3,5].
It can be shown that 6 swaps is the minimum swaps required to make a valid array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [9]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The array is already valid, so we return 0.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Greedy; Array
class Solution {
public:
    int minimumSwaps(vector<int>& nums) {
        int n = nums.size();
        int i = 0, j = 0;
        for (int k = 0; k < n; ++k) {
            if (nums[k] < nums[i] || (nums[k] == nums[i] && k < i)) {
                i = k;
            }
            if (nums[k] > nums[j] || (nums[k] == nums[j] && k > j)) {
                j = k;
            }
        }
        if (i == j) {
            return 0;
        }
        return i + n - 1 - j - (i > j);
    }
};