<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing order</strong>, remove some duplicates <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank"><strong>in-place</strong></a> such that each unique element appears <strong>at most twice</strong>. The <strong>relative order</strong> of the elements should be kept the <strong>same</strong>.</p>

<p>Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the <strong>first part</strong> of the array <code>nums</code>. More formally, if there are <code>k</code> elements after removing the duplicates, then the first <code>k</code> elements of <code>nums</code>&nbsp;should hold the final result. It does not matter what you leave beyond the first&nbsp;<code>k</code>&nbsp;elements.</p>

<p>Return <code>k</code><em> after placing the final result in the first </em><code>k</code><em> slots of </em><code>nums</code>.</p>

<p>Do <strong>not</strong> allocate extra space for another array. You must do this by <strong>modifying the input array <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in-place</a></strong> with O(1) extra memory.</p>

<p><strong>Custom Judge:</strong></p>

<p>The judge will test your solution with the following code:</p>

<pre>
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 &lt; k; i++) {
    assert nums[i] == expectedNums[i];
}
</pre>

<p>If all assertions pass, then your solution will be <strong>accepted</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,2,2,3]
<strong>Output:</strong> 5, nums = [1,1,2,2,3,_]
<strong>Explanation:</strong> 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).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,1,1,1,1,2,3,3]
<strong>Output:</strong> 7, nums = [0,0,1,1,2,3,3,_,_]
<strong>Explanation:</strong> 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).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>nums</code> is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        i = 1
        j = 1
        cnt = 1
        while i < len(nums):
            if nums[i] == nums[i-1]:
                cnt += 1
                if cnt > 2:
                    i += 1
                    continue
            else:
                cnt = 1
            nums[j] = nums[i]
            i += 1
            j += 1
        nums = nums[:j]
        return len(nums)
16200