# A solution to LeetCode Problem 26. Remove Duplicates from the Sorted Array in JavaScript

### Problem Statement:

Given an integer array `nums`

sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array `nums`

. More formally, if there are `k`

elements after removing the duplicates, then the first `k`

elements of `nums`

should hold the final result. It does not matter what you leave beyond the first `k`

elements.

Return `k`

* after placing the final result in the first *`k`

* slots of *`nums`

.

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

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array int[] expectedNums = [...]; // The expected answer with correct length int k = removeDuplicates(nums); // Calls your implementation assert k == expectedNums.length; for (int i = 0; i < k; i++) { assert nums[i] == expectedNums[i]; }

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,2] Output: 2, nums = [1,2,_] Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).

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 k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

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

`-100 <= nums[i] <= 100`

`nums`

is sorted in non-decreasing order.

### Solution:

```
function removeDuplicates(nums) {
let i = 0;
while (i < nums.length - 1) {
if (nums[i] === nums[i + 1]) {
nums.splice(i, 1);
} else {
i++;
}
}
return nums.length;
}
```

#### This solution has a time complexity of O(n) because we are looping through the array once and only making a constant number of splice calls. The space complexity is O(1) because we are not creating any new arrays or objects and only modifying the input array in place.

Alternatively, we can use the Set object to remove duplicates from the array. The Set object stores only unique values and automatically removes duplicates. We can then use the spread operator (…) to convert the Set back into an array.

Here is an example of this approach: