This is a **Leet Code** problem solved in Java, the problem description is:

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:**

As the input Array length is predetermined, it cannot be changed. Hence, the problem statement says ‘It doesn’t matter what values are set beyond the returned length.’. This means in an input array {1,1,2}, after removing the duplicates the return array could be {1,2,2} which is accepted. The number of unique integer is 2 which means the system will only check the first 2 elements of the array {1,2} and the third element is ignored for the test.

**Note**

- New List cannot be created, so we will modify the input array.
- No inbuilt functions are used.
- This is the simplest and one of the fastest solution. Here, the length of the unique elements is stored in int y.

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

### Like this:

Like Loading...

*Related*