Remove duplicates from sorted Array

 26. Remove Duplicates from Sorted Array

Easy

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 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).

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

CODE:

class Solution:

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

k=1 #nums cannot be less than 1 as per constraints

for i in range(1,len(nums)):

if nums[i]!=nums[i-1]:

nums[k]=nums[i]

k+=1

return k

OTHER CODE:

✅ Method 1: sort in place using [:]
	def removeDuplicates(self, nums: List[int]) -> int:
		nums[:] = sorted(set(nums))
		return len(nums)
Time Complexity: O(n)
Space Complexity: O(1)
❌ Common Wrong Answers:
nums = sorted(set(nums))
	return len(nums)
nums =  doesn't replace elements in the original list.
nums[:] = replaces element in place
In short, without [:], we're creating a new list object, which is against what this problem is asking for:
"Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory."

Method 3: Using .pop()
	def removeDuplicates(self, nums: List[int]) -> int:
		i = 1
		while i < len(nums):
			if nums[i] == nums[i - 1]:
				nums.pop(i)
			else:
				i += 1
		return len(nums)
Method 4: Using OrderedDict.fromkeys()
#ordereddict preserves order ,could be done by dict.fromkeys(nums) also
#this stores nums integers as keys and as keys are unique in dict the length 
is of unique elemnents
from collections import OrderedDict
class Solution(object):
    def removeDuplicates(self, nums):
        nums[:] =  OrderedDict.fromkeys(nums)
        return len(nums)