## Description

https://leetcode.com/problems/remove-duplicates-from-sorted-array/

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:**

Givennums=[1,1,2], Your function should return length =, with the first two elements of`2`

being`nums`

and`1`

respectively. It doesn't matter what you leave beyond the returned length.`2`

**Example 2:**

Givennums=[0,0,1,1,1,2,2,3,3,4], Your function should return length =, with the first five elements of`5`

being modified to`nums`

,`0`

,`1`

,`2`

, and`3`

respectively. It doesn't matter what values are set beyond the returned length.`4`

**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:

//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]); }

## Explanation

Two pointers, a slow pointer points to the first non-duplicate element, and a faster pointer keeps moving to the end of the array, if a non-duplicate number is found, copy the value to the slow pointer’s next position.

## Java Solution

class Solution { public int removeDuplicates(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int i = 0; for (int j = 0; j < nums.length; j++) { if (nums[j] != nums[i]) { i++; nums[i] = nums[j]; } } return i + 1; // Time Complextiy: O(n) // Space Complexity: O(1) } }

## Python Solution

```
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
if not nums:
return 0
i = 0
j = 0
while j < len(nums):
if nums[j] == nums[i]:
j += 1
else:
i += 1
nums[i] = nums[j]
return i + 1
```

- Time complexity: O(N)
- Space complexity: O(1)

good!