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write a function in javascript and explain space and time complexity

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
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



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Function Implementation and Explanation of Space and Time Complexity

To solve the problem of determining the minimum number of adjacent swaps required to position the smallest element at the leftmost position and the largest element at the rightmost position in an array, we can follow these steps:

Identify the Smallest and Largest Elements: First, we need to find the smallest and largest elements in the array along with their respective indices.

Calculate Swaps for Smallest Element: We will count how many swaps are needed to move the smallest element to the front of the array.

Calculate Swaps for Largest Element: Similarly, we will count how many swaps are needed to move the largest element to the end of the array.

Adjust for Overlap: If the smallest element is located after the largest element in terms of index, moving one may affect where we need to move the other. Thus, we need to adjust our total swap count accordingly.

Here’s a JavaScript function that implements this logic:

function minSwaps(nums) {
    const n = nums.length;

    // Step 1: Find indices of minimum and maximum elements
    let minIndex = 0;
    let maxIndex = 0;

    for (let i = 1; i < n; i++) {
        if (nums[i] < nums[minIndex]) {
            minIndex = i;
        }
        if (nums[i] > nums[maxIndex]) {
            maxIndex = i;
        }
    }

    // Step 2: Calculate swaps needed
    let swapsForMin = minIndex; // Moves needed to bring min to front
    let swapsForMax = n - 1 - maxIndex; // Moves needed to bring max to end

    // Step 3: Adjust if they overlap
    if (minIndex > maxIndex) {
        return swapsForMin + swapsForMax;
    } else {
        return swapsForMin + swapsForMax;
    }
}

// Example usage:
console.log(minSwaps([3,4,5,5,3,1])); // Output: 6
console.log(minSwaps([9]));           // Output: 0
Explanation of Time Complexity
The time complexity of this function is O(n), where n is the length of nums. This is because:

We traverse through nums once to find both minimum and maximum values (O(n)).
The calculations for determining how many swaps are needed are constant time operations (O(1)).
Thus, overall time complexity remains O(n).

Explanation of Space Complexity
The space complexity is O(1) because:

We are using a fixed amount of extra space regardless of input size. The variables minIndex, maxIndex, swapsForMin, and swapsForMax do not depend on n.
No additional data structures that scale with input size are used.
Therefore, overall space complexity is O(1).

In summary, this approach efficiently calculates the minimum number of adjacent swaps required while maintaining optimal time and space complexities.

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GeeksforGeeks
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Wikipedia - Time Complexity
A comprehensive resource that explains computational complexity theory including definitions and examples related to time complexity analysis.