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# Remove Duplicates from Sorted Array

## [Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array)

Given a sorted array *nums*, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) 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**](https://en.wikipedia.org/wiki/In-place_algorithm) 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:

```

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
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.

## Solutions

### 🧠 Cpp

```cpp
#include <algorithm>
class Solution {
public:
    int removeDuplicates(vector<int>& nums)
    {
        nums.erase(std::unique(nums.begin(), nums.end()), nums.end() );
        /*
        for (auto prev_value = nums.begin(), i = prev_value+1;
            i < nums.end();)
        {
            if(*i == *prev_value)
                nums.erase(i);
            else
            {
                prev_value = i;
                i++;
            }
        }
        */
        return nums.size();
    }
};
```