## Question

Given an integer array `nums`

sorted in **non-decreasing order**, remove some duplicates **in-place** such that each unique element appears **at most twice**. 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 = [1,1,1,2,2,3]Output:5, nums = [1,1,2,2,3,_]Explanation:Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).

**Example 2:**

Input:nums = [0,0,1,1,1,1,2,3,3]Output:7, nums = [0,0,1,1,2,3,3,_,_]Explanation:Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).

**Constraints:**

`1 <= nums.length <= 3 * 10`

^{4}`-10`

^{4}<= nums[i] <= 10^{4}`nums`

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

## Python Solution

class Solution: def removeDuplicates(self, A: List[int]) -> int: p=1 if A==[]: return d={A[0]:1} for i in range(1,len(A)): if A[i] in d: d[A[i]] += 1 else: d[A[i]] = 1 if d[A[i]]<=2: A[p]=A[i] p+=1 return p