Remove Duplicates From Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.
Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by >the caller.
// using the length returned by your function, it prints the >first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

Solution:

The requirement here is to remove all the duplicates from the array such that each element occurs only once.

To solve this problem, we can use two-pointer approach here. We will have a fast pointer called 'i' and a slow pointer called 'j'. We will compare each element with the element a location 'j'. If we encounter an element which is greater than the element at 'j' then we will add that element to position 'j+1' and update 'j'.

Github Link Link

Blog Link Link

Time Complexity: O(n)

Space Complexity: O(1)