# Remove Duplicates from Sorted Array | Leetcode #26 | Easy

In this post I will discuss the solution to the leetcode problem — Remove Duplicates from Sorted Array

## Problem:

Given a sorted array *nums*, remove the duplicates **in-place** such that each element appears only *once* and returns the new length.

Do not allocate extra space for another array, you must do this by **modifying the input array ****in-place** with O(1) extra memory.

**Clarification:**

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by **reference**, which means a modification to the input array will be known to the caller as well.

Internally you can think of this:

//numsis passed in by reference. (i.e., without making a copy)

int len = removeDuplicates(nums);// any modification tonumsin your function would be known by the caller.

// using the length returned by your function, it prints the firstlenelements.

for (int i = 0; i < len; i++) {

print(nums[i]);

}

**Example 1:**

**Input:** nums = [1,1,2]

**Output:** 2, nums = [1,2]

**Explanation:** Your function should return length = **2**, with the first two elements of *nums* being **1** and **2** respectively. It doesn't matter what you leave beyond the returned length.

**Example 2:**

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

**Output:** 5, nums = [0,1,2,3,4]

**Explanation:** Your function should return length = **5**, with the first five elements of *nums* being modified to **0**, **1**, **2**, **3**, and **4** respectively. It doesn't matter what values are set beyond the returned length.

**Constraints:**

`0 <= nums.length <= 3 * 104`

`-104 <= nums[i] <= 104`

`nums`

is sorted in ascending order.

## Solution:

There are 2 key pieces of information in the problem statement.

- The input array is sorted
- We can not allocate extra space.

If the problem allowed to have another array for storing output, then it would have been a piece of cake and you could solve it with just one iteration. But doing it in space is a really nice catch. Let’s jump to the solution then.

When we say “doing it in space”, we need to modify existing input array such that it does not have duplicates anymore. That means we need to iterate over the array at least once for sure. In worst case scenario, the array does not have any duplicates so we will have to return the length of the input array as output.

What we can do when the array has duplicates is — keep moving forward in the array as long as the current element and the previous elements are same otherwise store the next element at the current index. Sounds confusing? Let’s look at the steps.

- As a base condition, if the
`nums`

array is empty, we can straightaway return 0 as output. - We can iterate the array with 2 pointers. Let’s call them
`slow`

and`fast`

and initialize both of them to 1 as we will start the array iteration from the second element of the array. - While iterating the input array, we compare current element with the previous element. Since we are starting with index 1, there is no problem of running out of bound.
- If they are not equal, we store the element at index fast to the index of slow and increment slow by one. Otherwise slow remains the same.
- Return slow as the output as that is the count of distinct elements in the array.

Here is how the code looks like —

`class Solution {`

public int removeDuplicates(int[] nums) {

if (nums.length == 0) {

return 0;

}

int slow = 1;

for (int fast = 1; fast < nums.length; fast++) {

if (nums[fast] != nums[fast - 1]) {

nums[slow] = nums[fast];

slow++;

}

}

return slow;

}

}

Hope this helps! Happy coding! 🙂

If you think the solution can be improved or misses something, feel free to comment. There is always some room for improvement.

Find the solutions to the leetcode problems here — https://github.com/rishikeshdhokare/leetcode-problems