Remove Duplicates from Sorted Array

Explanation & Solution

Description

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. Return the number of unique elements.

More formally, if there are k unique elements, the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Input:nums = [1, 1, 2]

Output: 2

Explanation: The function returns k = 2, with the first two elements of nums being [1, 2].

Constraints

1 <= nums.length <= 3 * 10^4
-100 <= nums[i] <= 100
nums is sorted in non-decreasing order

Approach

Two Pointers pattern

1. Handle Empty Array

If nums.length === 0, return 0
There are no elements, so zero unique values

2. Initialize Two Pointers

Set i = 0 (slow pointer, tracks the position of the last unique element)
The loop variable j starts at 1 (fast pointer, scans through the array)

3. Scan With the Fast Pointer

Iterate j from 1 to nums.length - 1
For each element nums[j], compare it to nums[i]

4. Place Unique Elements

If nums[j] !== nums[i]:
Increment i
Copy nums[j] into nums[i]
This overwrites duplicate positions with the next unique value

5. Return the Count

After the loop, i is the index of the last unique element
Return i + 1 as the total number of unique elements

🧠 Key Insight

Because the array is already sorted, all duplicates are adjacent
The slow pointer i only advances when a new unique value is found
The fast pointer j skips over duplicates naturally

Time

O(n)single pass through the array

Space

O(1)in-place modification, no extra storage

Visualization

Input:

[1, 1, 2]

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Press ▶ or use ← → to step through

Pointer iPointer jProcessedDone

5 steps

Solution Code

Valid Palindrome

Move Zeroes