Explain the process of removing duplicates from a sorted array.

Explanation:

To remove duplicates from a sorted array, we can use a two-pointer technique. This approach efficiently processes the array in-place, making it suitable for space-constrained environments. Since the array is sorted, duplicates will occur consecutively, allowing us to easily identify and skip them.

Key Talking Points:

• Utilize the two-pointer technique to maintain a compact array of unique elements.
• In-place modification minimizes additional space usage.
• The sorted nature of the array helps in identifying duplicates efficiently.

NOTES:

Reference Table:

Pseudocode:

   def remove_duplicates(nums):
       if not nums:
           return 0
       
       unique_index = 0
       
       for i in range(1, len(nums)):
           if nums[i] != nums[unique_index]:
               unique_index += 1
               nums[unique_index] = nums[i]
       
       return unique_index + 1

• This function returns the length of the array containing unique elements after duplicates are removed in-place.

Follow-Up Questions and Answers:

1. Question: How would you handle removing duplicates if the array were not sorted?

   Answer: If the array is not sorted, you could use a hash set to track seen elements. By iterating through the array and adding unseen elements to the set, you can construct an array of unique elements. This method requires additional space proportional to the number of unique elements.

2. Question: Can you modify the solution to work with a stream of numbers where the entire array is not available at once?

   Answer: For a stream of numbers, you can use a data structure like a hash set to keep track of numbers that have already been seen. As each number arrives in the stream, check if it is in the set. If not, add it to the set and process it further. This way, you maintain a collection of unique numbers seen so far.

3. Question: What if the array is sorted in descending order? Would your approach change?

   Answer: The two-pointer technique would still work, but you'd need to adjust the comparison logic to account for the descending order. Essentially, you would look for changes in the reverse pattern, comparing current elements to the previous ones to detect duplicates.