Remove Duplicates from Sorted Array Two Pointer in DSA Python

Mental Model

Use two pointers to overwrite duplicates in a sorted array, keeping only unique elements at the front.

Analogy: Imagine sorting your books on a shelf and removing extra copies by sliding unique books forward, ignoring repeats behind.

Array: [1, 1, 2, 2, 3, 4, 4]
Pointers: i ↑ j ↑
Initially both at start, i marks last unique, j scans ahead

Dry Run Walkthrough

Input: array: [1, 1, 2, 2, 3]

Goal: Remove duplicates so array starts with unique elements only

Step 1: Start with i=0 (first unique), j=1 scanning next element

[1 (i), 1 (j), 2, 2, 3]

Why: We compare j to i to find new unique elements

Step 2: j=1 equals i=0 value (1), so skip duplicate, move j forward

[1 (i), 1, 2 (j), 2, 3]

Why: Duplicate found, do not move i, just advance j

Step 3: j=2 value (2) differs from i=0 value (1), increment i and copy j's value to i

[1, 2 (i), 2 (j), 2, 3]

Why: Found new unique, move i forward and update array

Step 4: j=3 value (2) equals i=1 value (2), skip duplicate, move j forward

[1, 2 (i), 2, 2 (j), 3]

Why: Duplicate again, only advance j

Step 5: j=4 value (3) differs from i=1 value (2), increment i and copy j's value

[1, 2, 3 (i), 2, 3 (j)]

Why: New unique found, move i and update array

Result:

Final array start: [1, 2, 3] with length 3

Annotated Code

DSA Python

from typing import List

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        if not nums:
            return 0
        i = 0  # last unique element index
        for j in range(1, len(nums)):
            if nums[j] != nums[i]:
                i += 1
                nums[i] = nums[j]  # overwrite duplicate
        return i + 1

# Driver code
nums = [1, 1, 2, 2, 3]
sol = Solution()
length = sol.removeDuplicates(nums)
print("Unique length:", length)
print("Array after removing duplicates:", nums[:length])

if not nums:

handle empty array edge case

i = 0 # last unique element index

initialize pointer i to track last unique element

for j in range(1, len(nums)):

j scans through array from second element

if nums[j] != nums[i]:

check if current element is new unique

i += 1

move i forward to next unique position

nums[i] = nums[j] # overwrite duplicate

overwrite duplicate with new unique element

return i + 1

return count of unique elements

Output

Unique length: 3

Array after removing duplicates: [1, 2, 3]

Complexity Analysis

Time: O(n) because we scan the array once with pointer j

Space: O(1) because we modify the array in place without extra space

vs Alternative: Compared to creating a new array for uniques (O(n) space), this uses constant space by overwriting duplicates

Edge Cases

empty array

returns 0 immediately

DSA Python

if not nums:

array with all duplicates

returns 1 with first element only

DSA Python

if nums[j] != nums[i]:

array with no duplicates

returns original length, array unchanged

DSA Python

if nums[j] != nums[i]:

When to Use This Pattern

When you see a sorted array and need to remove duplicates in place, use two pointers to overwrite duplicates efficiently.

Common Mistakes

Mistake: Starting i pointer at 1 instead of 0 causes index errors or misses first element
Fix: Initialize i to 0 to mark the first unique element

Mistake: Copying nums[j] to nums[i] without incrementing i leads to overwriting same position
Fix: Increment i before copying to ensure unique elements are placed correctly

Summary

Removes duplicates from a sorted array by overwriting duplicates with unique elements using two pointers.

Use when you want to remove duplicates in place without extra space.

The key is to keep one pointer for the last unique element and another to scan ahead for new uniques.