# 2340 Minimum Adjacent Swaps to Make a Valid Array

You are given a 0-indexed integer array `nums`.

Swaps of adjacent elements are able to be performed on `nums`.

A valid array meets the following conditions:

• The largest element (any of the largest elements if there are multiple) is at the rightmost position in the array.
• The smallest element (any of the smallest elements if there are multiple) is at the leftmost position in the array.

Return the minimum swaps required to make `nums` a valid array.

Example 1:

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

Example 2:

``````Input: nums = [9]
Output: 0
Explanation: The array is already valid, so we return 0.
``````

Constraints:

• `1 <= nums.length <= 105`
• `1 <= nums[i] <= 105`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 `````` ``````class Solution: def minimumSwaps(self, nums: List[int]) -> int: min_val, min_idx = math.inf, -1 max_val, max_idx = 0, -1 for i, num in enumerate(nums): if num < min_val: min_val, min_idx = num, i if num >= max_val: max_val, max_idx = num, i return ( min_idx + (len(nums) - 1 - max_idx) - (1 if min_idx > max_idx else 0) )``````