Remove Duplicates from Sorted Array LeetCode Solution…

Approach Summary

Slow pointer is the write head. Fast pointer scans; write only when a new unique element is found.

Practice Mode

Solve it yourself before revealing the code

DSA interviews test reasoning from brute force to optimal. Think through the constraints first.

Full Solution & Approach

The array is sorted, so every duplicate of a value sits directly next to the original. Use a write pointer that tracks where the next unique element should be placed and a read pointer that scans the array. For every element, if it differs from the previous element written, copy it to the write position and advance the write pointer; otherwise skip it. Since the input is sorted, "different from the previous written element" is exactly the condition for being a new unique value. At the end the first k positions hold the unique elements in order and k is the answer. This is the in-place filter pattern: a read pointer always advances, a write pointer only advances when something should be kept. The same skeleton removes zeroes, removes elements by value, and compresses in place.

A single pass where each element is read once and written at most once — O(n) time. No extra arrays — O(1) space.

Solution Code

Solution

public int removeDuplicates(int[] nums) {
    int write = 0;
    for (int read = 0; read < nums.length; read++) {
        if (write == 0 || nums[read] != nums[write - 1]) {
            nums[write] = nums[read];
            write++;
        }
    }
    return write;
}

int removeDuplicates(vector<int>& nums) {
    int write = 0;
    for (int read = 0; read < (int)nums.size(); read++) {
        if (write == 0 || nums[read] != nums[write - 1]) {
            nums[write] = nums[read];
            write++;
        }
    }
    return write;
}

def remove_duplicates(nums: list[int]) -> int:
    write = 0
    for read in range(len(nums)):
        if write == 0 or nums[read] != nums[write - 1]:
            nums[write] = nums[read]
            write += 1
    return write

function removeDuplicates(nums) {
  let write = 0;
  for (let read = 0; read < nums.length; read++) {
    if (write === 0 || nums[read] !== nums[write - 1]) {
      nums[write] = nums[read];
      write++;
    }
  }
  return write;
}

Edge Cases to Watch

Empty array — return 0
All elements identical — k is 1
No duplicates — k equals the array length and no writes shift anything
Single element — return 1

Algorithmic Pattern

Master the Two Pointers Pattern

View Pattern Guide →

Study the core invariant, step-by-step decision framework, and common interview traps for the Two Pointers pattern.

Framework: Two Pointers • Interview Prep: Tested at FAANG • Explore all Two Pointers problems & invariants →

How to Recognize This Pattern

In-place deduplication of sorted array

Complexity Analysis

Time Complexity

O(n)

Space Complexity

O(1)

Frequently Asked Questions

What is the optimal time and space complexity for Remove Duplicates from Sorted Array?

The optimal solution for Remove Duplicates from Sorted Array achieves O(n) runtime complexity and requires O(1) auxiliary space. A single pass where each element is read once and written at most once — O(n) time. No extra arrays — O(1) space.

How do I identify that Remove Duplicates from Sorted Array requires the Two Pointers pattern?

In technical interviews, look for key problem signals: In-place deduplication of sorted array. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Two Pointers strategy.

What critical edge cases must be handled in Remove Duplicates from Sorted Array?

Essential edge cases to verify include: Empty array — return 0; All elements identical — k is 1; No duplicates — k equals the array length and no writes shift anything; Single element — return 1. Always walk through these scenarios with your interviewer before coding.

Which tech companies frequently ask Remove Duplicates from Sorted Array?

Remove Duplicates from Sorted Array is a Easy-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Apple, Microsoft.

Remove Duplicates from Sorted Array LeetCode Solution