Video #3: LeetCode 26 Remove Duplicates from Sorted Array | Java DSA Solution

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Video #3: LeetCode 26 Remove Duplicates from Sorted Array | Java DSA Solution

In this video, we solve Video #3 of our LeetCode series: LeetCode 26 - Remove Duplicates from Sorted Array using Java.

This is a very important beginner-friendly DSA problem based on arrays and the two-pointer technique. In this problem, we are given a sorted array, and we need to remove duplicate elements in-place so that each unique element appears only once.

The main condition of this problem is that we should not use extra space for another array. We have to modify the given array itself and return the number of unique elements.

In this solution, we use a simple two-pointer approach. Since the array is already sorted, duplicate elements will always appear next to each other. We keep one pointer to track the position where the next unique element should be placed, and another pointer to scan the array.

Whenever we find a new unique element, we place it at the correct position and increase the count of unique elements. At the end, we return the value of k, which represents the total number of unique elements in the array.

Example:

Input:
nums = [1, 1, 2]

Output:
2

Modified array:
[1, 2, _]

Explanation:
There are 2 unique elements: 1 and 2.

Another Example:

Input:
nums = [0,0,1,1,1,2,2,3,3,4]

Output:
5

Modified array:
[0,1,2,3,4,_,_,_,_,_]

In this video, you will learn:
✅ What is LeetCode 26 Remove Duplicates from Sorted Array?
✅ How to solve array problems step by step
✅ Why sorted array helps in removing duplicates
✅ How the two-pointer approach works
✅ How to modify the array in-place
✅ Java solution explanation
✅ Dry run with examples
✅ Time and space complexity
✅ Beginner-friendly DSA logic

Approach Used:
We start by assuming the first element is already unique, so k starts from 1. Then we loop through the array from index 1. If the current element is different from the previous element, it means we found a new unique value. We place that value at index k and increase k. Finally, k gives us the count of unique elements.

Time Complexity:
O(n), because we visit each element of the array only once.

Space Complexity:
O(1), because we solve the problem in-place without using any extra array.

This problem is very useful for coding interview preparation because it teaches array traversal, in-place modification, and the two-pointer pattern. If you are learning DSA in Java or preparing for placements, this problem is a must-solve.

Watch the full video to understand the logic clearly and try to dry run the example yourself for better practice.

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Видео Video #3: LeetCode 26 Remove Duplicates from Sorted Array | Java DSA Solution канала Rv Ranjan - Tech