# 2340. Minimum Adjacent Swaps to Make a Valid Array

## Description

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

## Solutions

• class Solution {
public int minimumSwaps(int[] nums) {
int n = nums.length;
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 ? 1 : 0);
}
}

• 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);
}
};

• class Solution:
def minimumSwaps(self, nums: List[int]) -> int:
i = j = 0
for k, v in enumerate(nums):
if v < nums[i] or (v == nums[i] and k < i):
i = k
if v >= nums[j] or (v == nums[j] and k > j):
j = k
return 0 if i == j else i + len(nums) - 1 - j - (i > j)

• func minimumSwaps(nums []int) int {
var i, j int
for k, v := range nums {
if v < nums[i] || (v == nums[i] && k < i) {
i = k
}
if v > nums[j] || (v == nums[j] && k > j) {
j = k
}
}
if i == j {
return 0
}
if i < j {
return i + len(nums) - 1 - j
}
return i + len(nums) - 2 - j
}

• function minimumSwaps(nums: number[]): number {
let i = 0;
let j = 0;
const n = nums.length;
for (let 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;
}
}
return i == j ? 0 : i + n - 1 - j - (i > j ? 1 : 0);
}