26. Remove Duplicates from Sorted Array

Easy

Given an integer arraynums sorted innon-decreasing order, remove the duplicatesin-place such that each unique element appears onlyonce. Therelative order of the elements should be kept thesame.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in thefirst part of the arraynums. More formally, if there arek elements after removing the duplicates, then the firstk elements ofnums should hold the final result. It does not matter what you leave beyond the first k elements.

Returnk after placing the final result in the firstk slots ofnums.

Donot allocate extra space for another array. You must do this bymodifying the input arrayin-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input arrayint[] expectedNums = [...]; // The expected answer with correct lengthint k = removeDuplicates(nums); // Calls your implementationassert k == expectedNums.length;for (int i = 0; i < k; i++) {    assert nums[i] == expectedNums[i];}

If all assertions pass, then your solution will beaccepted.

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 innon-decreasing order.

Python Solution Using an Iterative approach

class Solution:    def searchInsert(self, nums: List[int], target: int) -> int:        for i in range(len(nums)):            if nums[i] >= target:                return i                break            elif nums[i] > target and target < nums[i + 1]:                return i+1                break            elif target > nums[-1]:                return len(nums)            elif target < nums[0]:                return 0                break        return i

Time Complexity: O(N)
Auxiliary Space: O(1)

A better alternative -- Python solution using Binary search.

class Solution:    def searchInsert(self, nums: List[int], target: int) -> int:        n = len(nums)        start = 0        end = n - 1        while start <= end:            mid = (start + end) // 2            if nums[mid] == target:                return mid            elif nums[mid] < target:                start = mid + 1            else:                end = mid - 1        return end + 1

Time Complexity: O(log N)
Auxiliary Space: O(1)