leetcode - 26 Remove the duplicate from sorted array - inplace

 

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Question:

26. Remove Duplicates from Sorted Array

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.

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).

 

Solution:

Here we using two pointer slow and fast solution

we start slow poinet at position 0 

and  start second fast pointer traverse array from 1 to  end

and condition to  check if  the  current  iteration is unique and does not have match

and the n we increment the slow pointer to  next  element  and  

update the  elements position with   fast pointer iteration element

in last  we delete all the  element  from to  end/start

For loop do this:

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

the line del nums[slow+1:]

will eliminate  the  rest element

class Solution:

    def removeDuplicates(self, nums: List[int]) -> int:

        slow = 0

        for fast in range(1,len(nums)): # Start from the second element

            if nums[fast] !=nums[slow]:

                slow = slow + 1 # Move the slow pointer to the next unique position

                nums[slow] = nums[fast] # Place the unique element in the right position

        del nums[slow+1:]

        print(nums)