## Description

Given a sorted array *nums*, remove the duplicates **in-place** such that each element appear only *once* and return 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.

**Example 1:**

```
Given nums = [1,1,2],
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:**

```
Given nums = [0,0,1,1,1,2,2,3,3,4],
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.
```

**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 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]);
}
```

## Solutions

### 1. Array

```
# Time: O(n)
# Space: O(1)
# class Solution:
# def removeDuplicates(self, nums: List[int]) -> int:
# i = 0
# while i < len(nums):
# if i > 0 and nums[i-1] == nums[i]:
# nums = nums[:i] + nums[i+1:]
# else:
# i += 1
# return len(nums)
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
tail = 0
n = len(nums)
for i in range(1, n):
# Override
if nums[i] != nums[tail]:
tail += 1
nums[tail] = nums[i]
return tail + 1
# 161/161 cases passed (84 ms)
# Your runtime beats 77.76 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (14.3 MB)
```