In this tutorial, we will teach you How you can remove duplicates from sorted array. It’s very commonly asked question in DSA Interview. Also, we will show you Animation to understand the algorithm better way. his problem helps interviewers evaluate your understanding of arrays, in-place modification, and optimized logic using two pointers.

We’ll use C++, Java, Puthon , and JavaScript programming language , you can choose your favorite programming language. We’ll also analyze the time and space complexity, making it easier for you to remember and apply this technique in real interviews. you can solve it on Leetcode platform.

This problem helps to understand about you knowledge of :

• the array traversal skill.
• your logical thinking.
• your knowledge of time & space complexity.
• Optimized logic using two pointer.

Problem Statement

You can read the complete problem statement on the LeetCode platform by clicking here.

Optimized Algorithm – Two Pointer Technique – Explained using Animation

Let an Array nums is given.

Step-by-step Explanation:

1. Edge case
   • If the array is empty (nums.length === 0), return 0
2. Initialize a pointer
   • Let k = 1
   • This pointer represents the position of the next unique element
3. Traverse the array
   • Start from index 1 to n - 1
   • Compare current element nums[i] with previous element nums[i - 1]
4. If elements are different
   • It means we found a new unique element
   • Place it at index k
   • Increment k
5. Return k
   • k is the count of unique elements
   • First k elements of the array are unique and sorted

Dry run example

nums = [1, 1, 2]

Final array: [1, 2, _]
Output: k = 2

Remove Duplicates From Sorted Array – Using Animation

Solution Using C++, Java, Python, JavaScript

To solve this problem , we have used below programming language

• C++
• Java
• Python
• JavaScript

Each implementation follows the same algorithm, helping you understand how logic remains constant while syntax changes.

int removeDuplicates(vector<int>& nums) {
        if (nums.size() == 0)
            return 0;

        int k = 1; // pointer for unique elements

        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] != nums[i - 1]) {
                nums[k] = nums[i];
                k++;
            }
        }

        return k;
    }

public int removeDuplicates(int[] nums) {
        if (nums.length == 0)
            return 0;

        int k = 1; // pointer for unique elements

        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[i - 1]) {
                nums[k] = nums[i];
                k++;
            }
        }

        return k;
    }

def removeDuplicates(self, nums):
        if len(nums) == 0:
            return 0

        k = 1  # pointer for unique elements

        for i in range(1, len(nums)):
            if nums[i] != nums[i - 1]:
                nums[k] = nums[i]
                k += 1

        return k

var removeDuplicates = function(nums) {

    if ( nums.length === 0 )
    return 0;

    let k = 1; //pointer for unique elements

    for ( let i = 1; i < nums.length; i++ ) {
        if ( nums[i] !== nums[i-1] ) {
            nums[k] = nums[i];
            k++;
        }
    }

    return k;
    
};

Output:-

Conclusion:-

Thank you for reading this tutorial. Our motivation is to share DSA concepts and problem-solving techniques in a simple and easy-to-understand way. We hope this explanation helped you clearly understand the Remove Duplicates From Sorted Array and its optimized solution.

We encourage you to explore more DSA problems and interview questions on our platform to build strong fundamentals and gain confidence for technical interviews.

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