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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. Then return the number of unique elements in `nums`

.
Consider the number of unique elements of `nums`

to be `k`

, to get accepted, you need to do the following things:

Change the array `nums`

such that the first `k`

elements of `nums`

contain the unique elements in the order they were present in `nums`

initially. The remaining elements of `nums`

are not important as well as the size of `nums`

.
Return `k`

.s.

**Examples:**

- Input:
`nums = [1,1,2]`

, Output:`2`

. The resulting`nums`

will look lie:`nums = [1,2,_]`

- Input:
`nums = [0,0,1,1,1,2,2,3,3,4]`

, Output:`5`

. The resulting`nums`

will look lie:`nums = [0,1,2,3,4,_,_,_,_,_]`

One way to solve this problem is to use two pointers, a left and a right one. The left shows the end of the non-duplicated array (i.e., the result). The right shows the current element of the matrix. As we move the right index, we check whether it is the same as the left one. If it is, then it means we have a duplicate, and we can ignore that element: hence we can move the right index to the right. If not, then we have a new element, and hence we can copy the right element to the \(left+1\) place in the array, and move on to the next elemnt.

```
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
# use two pointers: left and right.
# the left tracks the non-duplicated array; the right tracks how many elements
# we have seen so far.
l,r = 0,1
n = len(nums)
while r < n:
if nums[r] == nums[l]:
# duplicate, move right index
r+=1
else:
l+=1
# replace the next element in nums[l] w the current in nums[r]
nums[l] = nums[r]
r+=1
return l+1
```

If we wanted to run it on an example:

`## [0, 1, 2, 3, 4]`

The time complexity of this solution is \(O(N)\), since there is no nested loop.

Array, Two pointers

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