Python 3.14 (Coding): Remove Duplicates Test

The solution to the problem implements a funtion which returns an integer representing the length of the modified vector after removing any number of duplicates. The function begins with a check to see if the size of the input vector nums is less than or equal to 1. If it is, then there are no duplicates to remove, and the function simply returns the size of the vector. If the size of nums is greater than 1, the function proceeds with the duplicate removal process. It initializes two variables: slow and fast. slow represents the index of the last non-duplicate element found so far, and fast is used to iterate through the vector to find duplicates. The function enters a loop that continues until fast reaches the end of the vector (n is the size of nums). Inside the loop, it checks if the current element (nums[fast]) is equal to the previous element (nums[fast - 1]). If they are equal, it means a duplicate is found. If a duplicate is found, the code enters another inner loop with the condition if (++fast >= n) break;. This loop increments fast until it finds the next non-duplicate element or reaches the end of the vector. This ensures that fast is pointing to the first occurrence of a non-duplicate element after the series of duplicates. After exiting the inner loop, the code checks if fast is still less than the size of the vector (fast < n). If it is, it means a non-duplicate element was found after the series of duplicates. In this case, it assigns the value of this non-duplicate element to nums[++slow], effectively moving the non-duplicate element to the next position in the vector. The outer loop continues, and the process repeats for the next set of duplicates or non-duplicate element. Finally, the function returns slow + 1, which represents the length of the modified vector after removing duplicates. Adding 1 is necessary because slow represents the index of the last non-duplicate element, not the length of the vector itself.