Problem 10: Duplicate Removal

Hey everyone! 👋

Today, we're tackling a list manipulation problem: Removing Duplicates while Maintaining Order.

The Problem

The goal is to write a function that removes duplicates from a list while maintaining the original order of elements.

Example:

• remove_duplicates([1, 2, 2, 3, 1, 4]) should return [1, 2, 3, 4]

The Solution

Here is the Python implementation:

def remove_duplicates(lst):
    """
    Removes duplicates from a list while maintaining order.
    """
    seen = set()
    result = []

    for i in lst:
        if i not in seen:
            seen.add(i)
            result.append(i)
    return result

# Test case
print(remove_duplicates([1, 2, 2, 3, 1, 4]))
# Output: [1, 2, 3, 4]

Code Breakdown

Let's walk through the code line by line:

1. def remove_duplicates(lst):

   • Defines a function named remove_duplicates that takes one parameter lst (a list).

2. seen = set()

   • Creates an empty set called seen to keep track of unique elements we've encountered. Using a set allows for efficient O(1) lookups.

3. result = []

   • Creates an empty list called result to store the final list without duplicates.

4. for i in lst:

   • Starts a loop that iterates through each element in the input list lst.

5. if i not in seen:

   • Checks if the current element i is not in the seen set (i.e., it's the first time we're seeing this element).

6. seen.add(i)

   • If the element is not in seen, adds it to the set to mark it as seen.

7. result.append(i)

   • Appends the element to the result list to maintain the order of first occurrence.

8. return result

   • Returns the final list with duplicates removed, in the order of their first appearance.

Example Walkthrough with [1, 2, 2, 3, 1, 4]

1. i = 1: Not in seen → add to seen and result

   • seen: {1}

   • result: [1]

2. i = 2: Not in seen → add to seen and result

   • seen: {1, 2}

   • result: [1, 2]

3. i = 2: Already in seen → skip

4. i = 3: Not in seen → add to seen and result

   • seen: {1, 2, 3}

   • result: [1, 2, 3]

5. i = 1: Already in seen → skip

6. i = 4: Not in seen → add to seen and result

   • seen: {1, 2, 3, 4}

   • result: [1, 2, 3, 4]

Final result: [1, 2, 3, 4]

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Happy coding! 💻