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### by Sumit Chourasia | Nov 01, 2020 | Category :coding | Tags : algorithmarraydata-structureleetcodemedium #### Remove Duplicates from Sorted Array II - Array - Medium - LeetCode

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice 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.

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 a modification to the input array will be known to the caller.

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]);
}

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3]
Explanation: Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. It doesn't matter what you leave beyond the returned length.
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 length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively. It doesn't matter what values are set beyond the returned length.

Constraints:

0 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
nums is sorted in ascending order.

``````public class Solution {
public int RemoveDuplicates(int[] nums) {
int index=0;
for(int i=0;i<nums.Length;i++){
if(index<2 || nums[i] > nums[index-2]){
nums[index++]=nums[i];
}
}

return index;
}
}``````

Time Complexity: O(n)

Space Complexity: O(1)