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**.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the **first part** of the array `nums`

. More formally, if there are `k`

elements after removing the duplicates, then the first `k`

elements of `nums`

should hold the final result. It does not matter what you leave beyond the first `k`

elements.

Return `k`

* after placing the final result in the first *`k`

* slots of *`nums`

.

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

**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:**

`0 <= nums.length <= 3 * 10`

^{4}`-10`

^{4}<= nums[i] <= 10^{4}`nums`

is sorted in**non-decreasing**order.

class Solution {
public int removeDuplicates(int[] nums) {
if(nums.length<=1){
return nums.length;
}
int i=1;
for(int j=1;j<nums.length;j++){
if(nums[j]!=nums[j-1]){
nums[i]=nums[j];
i++;
}
}
return i;
}
}

This question can easily be solved using two pointer approach, the first pointer denoting the distinct element while the second pointer will be traversing through all the elements of the array.

Time Complexity: O(n)

Space Complexity: O(1)