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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 * 10`

^{4}`-100 <= nums[i] <= 100`

`nums`

is sorted in**non-decreasing**order.

```
int removeDuplicates(vector<int>& nums) {
int i = 0;
for (int n : nums)
if (!i || n > nums[i-1])
nums[i++] = n;
return i;
}
```

```
public int removeDuplicates(int[] nums) {
int i = 0;
for (int n : nums)
if (i == 0 || n > nums[i-1])
nums[i++] = n;
return i;
}
```

```
x = 1
for i in range(len(nums)-1):
if(nums[i]!=nums[i+1]):
nums[x] = nums[i+1]
x+=1
return(x)
```

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