Remove Duplicates with O(1)

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Unit 4 Session 1 (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• How should the function behave with an empty list?
  • It should return 0 because there are no elements to process.
• Are there any constraints on the sorting of the list?
  • The list is assumed to be sorted, which allows for efficient duplicate removal.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a two-pointer technique to modify the list in place, maintaining the order and uniqueness of elements.

1) If the list is empty, return 0.
2) Initialize a pointer j to 1.
3) Iterate through the list from the second element:
  a) If the current element is different from the previous one, place it at the j-th position.
  b) Increment j.
4) Trim all elements after the j-th index.
5) Return j as the new length of the list.

⚠️ Common Mistakes

• Not handling the case of an empty list properly.
• Incorrectly moving the j pointer could lead to either missing unique elements or not removing all duplicates.

I-mplement

def remove_duplicates(nums):
    if not nums:
        return 0
    
    # Pointer j for the position of the next unique element
    j = 1
    
    # Iterate through the array with i
    for i in range(1, len(nums)):
        if nums[i] != nums[i-1]:
            nums[j] = nums[i]
            j += 1
     
    nums[:] = nums[:j]

            
    return j  # The new length after removing duplicates