Remove Duplicates from a Sorted Array: Techniques, Examples, and Efficient Solutions

What Does It Mean to Remove Duplicates from a Sorted Array?

When you're asked to delete duplicates from a sorted list, you're trying to clean out the data so that each number only appears once — and do it in-place, i.e., without creating any additional space. Since the list is sorted, all of the duplicates are clustered together. That makes it quicker and more efficient to locate and delete them.

This issue is a favorite in coding interviews since it challenges your knowledge of array manipulation, two-pointer solution, and algorithmic efficiency. It also indicates how well you can perform in-place operations, which are vital in real-life scenarios where memory consumption becomes an issue.

Why Is This Problem Important?

Removing duplicates serves to sanitize data, eliminate redundancy, and enhance performance. For instance:

• In search systems, duplicates would skew results.
• In analytics, duplicate entries can cause biased insights.
• In systems that are memory-limited, duplicates incur wasted storage.

Understanding how to eliminate duplicates in an efficient manner provides you with a firm foundation in array-based algorithms and enhances your technical interview problem-solving ability.

How Does Sorting Assist in Eliminating Duplicates?

An array that is sorted is your ally here. Because the elements are already sorted, repeated elements come one after another. Which means you can literally just iterate through the array, compare each element to the previous one, and move on from repeated values.

This enables us to find the solution in linear time, or O(n), with constant space, or O(1), by employing a technique called the two-pointer technique.

What's the Two-Pointer Technique and Why Is It Optimal?

The two-pointer technique is one of the most effective methods for solving this problem. Here's the general idea:

• You employ two variables — say, slow and fast.
• fast scans through the whole array to locate the next unique number.
• slow marks the location for the next unique value.

Each time nums[fast] differs from nums[slow], you've located a new unique number. You add one to slowand insert the new value to nums[slow].

This strategy alters the initial array in-place, doesn't use extra space, and is extremely efficient on big data sets.

How Does This Work in Practice? A Step-by-Step Example

Let's take a typical scenario:

You have the sorted array:
[1, 1, 2, 3, 3, 4, 5, 5]

• Begin with slow = 0 and fast = 1.
• Compare nums[fast] with nums[slow].
• If they are not equal, that means you've discovered a new unique value.
• Move slow by one step and update nums[slow] = nums[fast].
• Continue until fast hits the end.

When the loop ends, the array's initial part — indices 0 through slow — will have all distinct elements in sorted order. You return slow + 1as the number of distinct elements.

The resulting array will be like this initially:
[1, 2, 3, 4, 5, _, _, _]
(The underscores are for values that do not matter anymore.)

Python Code to Remove Duplicates In-Place

Here is how you can achieve this in Python:

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def remove_duplicates(nums):
    if not nums:
        return 0

    slow = 0
    for fast in range(1, len(nums)):
        if nums[fast] != nums[slow]:
            slow += 1
            nums[slow] = nums[fast]
    return slow + 1

# Example usage
nums = [1, 1, 2, 3, 3, 4, 5, 5]
length = remove_duplicates(nums)
print("Length:", length)
print("Updated Array:", nums[:length])

This will print:

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2
Length: 5  
Updated Array: [1, 2, 3, 4, 5]

What Happens If the Array Is Empty or Has One Element?

These are significant edge cases that your code should handle nicely.

• Empty array: There is no work to do. Just return 0.
• Single-element array: No duplicates can be formed. Return 1.

Always ensure that your function won't crash on edge cases — that's something interviewers look for.

Common Mistakes You Should Avoid

Following are some traps that can catch you off guard:

• Returning the wrong length: Keep in mind, the new length is slow + 1, not slow.
• Skipping the first element: You must begin comparing from the second element.
• Not changing the array in-place: If you're using additional memory (e.g., sets), you're not satisfying the in-place condition.
• Mixing up slow and fast: Be sure slowpoints to where you write, and fastis checking for distinct items.

Can You Use Sets or Dictionaries to Eliminate Dupes?

Yes, you can, but not in this problem. A set would easily remove duplicates, but:

• It wouldn't preserve order.
• It takes additional space — which is prohibited in this problem.
• It won't be counted as an in-place solution.

The two-pointer algorithm is the most interview-perfect and space-saving solution.

How Can You Check That Your Solution Is Correct?

After your function executes, it returns the number of unique values. You can slice the array from the beginning to that number and verify:

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unique_count = remove_duplicates(nums)
unique_values = nums[:unique_count]
print(unique_values)

This will give you the correct, deduplicated array — in order and with no extra memory used.

How Does Removing Duplicates Improve Performance?

When your data is clean:

• Searching becomes faster.
• Sorting or merging other arrays becomes simpler.
• Algorithms like binary search become more accurate.
• You reduce unnecessary memory operations.

In short, deduplication helps your overall algorithm performance, especially when handling large datasets.

Are There Advanced Versions of This Problem?

Yes, and they’re popular in interviews too.

• Allow duplicates at most twice: Modify the logic to allow each value up to two times.
• Unsorted array duplicates: Requires a set or sorting first, since values aren’t grouped.
• Remove duplicates from linked lists: A similar technique, but adapted to node pointers instead of array indexes.