Remove Duplicates from Sorted Array Two Pointer in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of removing duplicates using two pointers

What is the output array after removing duplicates from the sorted array using the two-pointer technique?

DSA Python

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

result = remove_duplicates([1,1,2,2,3,4,4,5])
print(result)

A [1, 2, 3, 4]

B [1, 1, 2, 3, 4, 5]

C [1, 2, 2, 3, 4, 5]

D [1, 2, 3, 4, 5]

💡 Hint

Focus on how the two pointers move and overwrite duplicates.

The two-pointer method moves through the array, overwriting duplicates by shifting unique elements forward. The returned slice contains only unique elements.

❓ Predict Output

intermediate

Length of array after removing duplicates

What is the length of the array after removing duplicates using the two-pointer method on this input?

DSA Python

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

length = remove_duplicates([0,0,1,1,1,2,2,3,3,4])
print(length)

A 4

B 5

C 7

D 6

💡 Hint

Count how many unique elements remain after processing.

The unique elements are [0,1,2,3,4], so length is 5.

🔧 Debug

advanced

Identify the error in two-pointer duplicate removal

What error will this code produce when removing duplicates from a sorted array?

DSA Python

def remove_duplicates(nums):
    i = 0
    for j in range(0, len(nums)):
        if nums[j] == nums[i]:
            i += 1
            nums[i] = nums[j]
    return nums[:i+1]

print(remove_duplicates([1,1,2,3,3]))

A IndexError

B TypeError

C [1, 1, 2, 3, 3]

D [1, 2, 3]

💡 Hint

Check the condition inside the loop and how i is incremented.

The condition is reversed; i increments on duplicates causing i to go out of bounds.

❓ Predict Output

advanced

Final array state after removing duplicates with extra elements

What is the final state of the array after removing duplicates from [2,2,2,3,3,4,5,5,6] using the two-pointer method?

DSA Python

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

result = remove_duplicates([2,2,2,3,3,4,5,5,6])
print(result)

A [2, 3, 3, 4, 5, 6]

B [2, 2, 3, 4, 5, 6]

C [2, 3, 4, 5, 6]

D [2, 3, 4, 5]

💡 Hint

Only unique elements remain in the returned slice.

Duplicates are overwritten, leaving unique elements [2,3,4,5,6].

🧠 Conceptual

expert

Why does the two-pointer method work for removing duplicates in sorted arrays?

Which explanation best describes why the two-pointer technique correctly removes duplicates from a sorted array?

A Because the array is sorted, duplicates are adjacent, allowing one pointer to track unique elements and the other to scan ahead, overwriting duplicates efficiently.

B Because the array is sorted, the pointers can jump over duplicates without checking each element, reducing time complexity to O(log n).

C Because the array is sorted, the two pointers swap elements to sort the array again while removing duplicates.

D Because the array is sorted, the two pointers create a new array with only unique elements by copying elements one by one.

💡 Hint

Think about adjacency of duplicates in sorted arrays.

Duplicates appear next to each other in sorted arrays, so one pointer can track the last unique element while the other scans forward to find new unique elements to overwrite duplicates.