# Leetcode Remove Duplicates from Sorted Array problem solution

Feb 26, 2023

In the Leetcode Remove Duplicates from Sorted Array problem solution 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:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

## Solution in C Programming

``````int removeDuplicates(int* nums, int numsSize){
int i=0;
int temp=0;
int x=1;

if(numsSize>0)
x=numsSize;
int tnums[x];
if(numsSize==1){
for(i=0;i<numsSize-1;i++){
if(nums[i]!=nums[i+1]){
tnums[temp]=nums[i];
tnums[temp+1]=nums[i+1];
temp+=1;
}
}
for(i=0;i<temp;i++){
nums[i]=tnums[i];
}
return 1;
}
else if(numsSize==0)
return 0;

for(i=0;i<numsSize-1;i++){
if(nums[i]!=nums[i+1]){
tnums[temp]=nums[i];
tnums[temp+1]=nums[i+1];
temp+=1;
}
else{
tnums[temp]=nums[i];
}
}
for(i=0;i<=temp;i++){
nums[i]=tnums[i];
}
return temp+1;
}``````

## Solution in C++ Programming

``````class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int n= nums.size();
if(n<2)
return n;
int j=0;
for(int i=1; i<n; i++){
if(nums[i]!=nums[j]){
// If elements at i & j not equal, increment j and copy
// nums[i] to nums[j] so as to concentrate all unique
// elements in ascending order at the initial indices of the vector
j++;
nums[j]=nums[i];

}
}
// j location is the final indice of the
// unique element subarray at the beginning of nums vector
return j+1;
}
};``````

## Solution in Java Programming

``````class Solution {
public int removeDuplicates(int[] nums) {
if(nums.length == 0)
{
return 0;
}
int j = 0;
for(int i = 1 ; i < nums.length ; i++){
if(nums[j]!=nums[i]){
j = j+1;
nums[j] = nums[i];
}
}
return j+1;
}
}``````

## Solution in Python Programming

``````class Solution(object):
def removeDuplicates(self, nums):
if len(nums) <= 1: return len(nums)
for i in range(len(nums) - 1, 0, -1):
if nums[i] == nums[i - 1]: nums.pop(i)
return len(nums)``````

## Solution in C# Programming

``````public class Solution {
public int RemoveDuplicates(int[] nums) {
var currentPosition = 0;
var currentVal = int.MinValue;

for (int i = 0; i < nums.Length; i++)
{
if (nums[i] > currentVal)
{
currentVal = nums[i];
nums[currentPosition] = currentVal;
currentPosition++;
}
}

return currentPosition;
}
}``````

#### By Neha Singhal

Hi, my name is Neha singhal a software engineer and coder by profession. I like to solve coding problems that give me the power to write posts for this site.