26. Remove Duplicates from Sorted Array : Leetcode

 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.

Consider the number of unique elements in nums to be k​​​​​​​​​​​​​​. After removing duplicates, return the number of unique elements k.

The first k elements of nums should contain the unique numbers in sorted order. The remaining elements beyond index k - 1 can be ignored.

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.

Solution 1: using while loop

class Solution:

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

if not nums:

return 0

l,r=0,1

while r< len(nums):

if nums[l] != nums[r]:

l+=1

nums[l]=nums[r]

r+=1

return l+1 # "extra" values are still there at the end of the array, but we just ignore them

Solution 2: using for loop

class Solution:

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

if not nums:

return 0

l=0

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

if nums[l] != nums[r]:

l+=1

nums[l]=nums[r]

r+=1

return l+1 # "extra" values are still there at the end of the array, but we just ignore them

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