# Remove Duplicates from Sorted Array — LeetCode

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.

**Constraints:**

`0 <= nums.length <= 3 * 104`

`-100 <= nums[i] <= 100`

`nums`

is sorted in**non-decreasing**order

**Custom Judge:**

The judge will test your solution with the following code:

int[] nums = [...]; // Input array

int[] 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 be **accepted**.

## My Solution —

**class solution** {

**public: **

**int removeDuplicates**(vector<**int**>& nums)//given parameter's vector

{

** unordered_set**<**int**> s; // taken a set as a variable

**for**(**int **i:nums) // mapped through all the elements of nums

s.**insert**(i); //inserted the non duplicate elements in s

nums.**clear**() // emptied the vector nums

**for**(**int **i:s)

nums.**push_back**(i) //appended the elements from s to nums

**sort**(nums.**begin**(), nums.**end**()) //sorted nums

**return **s.**size**()

}

}

Runtime:

20 msMemory Usage:

18.9 MB

## Other Solution —

**class** **Solution** {

**public**:

**int** **removeDuplicates**(vector<**int**>& nums)

{

**int** j = 0;

**int** n = nums.size();

**if**(n <= 1) **return** n; //return the size of nums if n<=1

**for**(**int** i = 1; i<n; i++)

{

**if**(nums[i] != nums[j])

{

j++;

nums[j] = nums[i]; //

}

}

**return** j+1;

}

};

Runtime:

16 msMemory Usage:

18.2 MB