Remove duplicates from sorted array

Given a sorted array, the task is to remove the duplicate elements from the array.

Examples:

Input  : arr[] = {2, 2, 2, 2, 2}
Output : arr[] = {2}
         new size = 1

Input  : arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5}
Output : arr[] = {1, 2, 3, 4, 5}
         new size = 5

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1: (Using extra space)

Create an auxiliary array temp[] to store unique elements.
Traverse input array and one by one copy unique elements of arr[] to temp[]. Also keep track of count of unique elements. Let this count be j.
Copy j elements from temp[] to arr[] and return j

C++

// Simple C++ program to remove duplicates 
#include<iostream> 
using namespace std; 
  
// Function to remove duplicate elements 
// This function returns new size of modified 
// array. 
int removeDuplicates(int arr[], int n) 
{ 
    // Return, if array is empty 
    // or contains a single element 
    if (n==0 || n==1) 
        return n; 
  
    int temp[n]; 
  
    // Start traversing elements 
    int j = 0; 
    for (int i=0; i<n-1; i++) 
  
        // If current element is not equal 
        // to next element then store that 
        // current element 
        if (arr[i] != arr[i+1]) 
            temp[j++] = arr[i]; 
  
    // Store the last element as whether 
    // it is unique or repeated, it hasn't 
    // stored previously 
    temp[j++] = arr[n-1]; 
  
    // Modify original array 
    for (int i=0; i<j; i++) 
        arr[i] = temp[i]; 
  
    return j; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    // removeDuplicates() returns new size of 
    // array. 
    n = removeDuplicates(arr, n); 
  
    // Print updated array 
    for (int i=0; i<n; i++) 
       cout << arr[i] << " "; 
  
    return 0; 
} 

Java

   
// simple java program to remove 
// duplicates 
  
class Main 
{ 
    // Function to remove duplicate elements 
    // This function returns new size of modified 
    // array. 
    static int removeDuplicates(int arr[], int n) 
    { 
        // Return, if array is empty 
        // or contains a single element 
        if (n==0 || n==1) 
            return n; 
       
        int[] temp = new int[n]; 
          
        // Start traversing elements 
        int j = 0; 
        for (int i=0; i<n-1; i++) 
            // If current element is not equal 
            // to next element then store that 
            // current element 
            if (arr[i] != arr[i+1]) 
                temp[j++] = arr[i]; 
          
        // Store the last element as whether 
        // it is unique or repeated, it hasn't 
        // stored previously 
        temp[j++] = arr[n-1];    
          
        // Modify original array 
        for (int i=0; i<j; i++) 
            arr[i] = temp[i]; 
       
        return j; 
    } 
      
    public static void main (String[] args)  
    { 
        int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5}; 
        int n = arr.length; 
          
        n = removeDuplicates(arr, n); 
   
        // Print updated array 
        for (int i=0; i<n; i++) 
           System.out.print(arr[i]+" "); 
    } 
} 

Python3

# Python3 program to 
# remove duplicates 
# Function to remove 
# duplicate elements 
  
# This function returns 
# new size of modified 
# array. 
def removeDuplicates(arr, n): 
  
    # Return, if array is  
    # empty or contains 
    # a single element 
    if n == 0 or n == 1: 
        return n 
  
    temp = list(range(n)) 
  
    # Start traversing elements 
    j = 0; 
    for i in range(0, n-1): 
  
        # If current element is 
        # not equal to next 
        # element then store that 
        # current element 
        if arr[i] != arr[i+1]: 
            temp[j] = arr[i] 
            j += 1
  
    # Store the last element 
    # as whether it is unique 
    # or repeated, it hasn't 
    # stored previously 
    temp[j] = arr[n-1] 
    j += 1
      
    # Modify original array 
    for i in range(0, j): 
        arr[i] = temp[i] 
  
    return j 
  
# Driver code 
arr = [1, 2, 2, 3, 4, 4, 4, 5, 5] 
n = len(arr) 
  
# removeDuplicates() returns 
# new size of array. 
n = removeDuplicates(arr, n) 
  
# Print updated array 
for i in range(n): 
    print ("%d"%(arr[i]), end = " ") 

C#

// Simple C# program to remove 
// duplicates 
using System; 
  
class GFG { 
      
    // Function to remove duplicate 
    // elements This function returns 
    // new size of modified array. 
    static int removeDuplicates(int []arr, int n) 
    { 
          
        // Return, if array is empty 
        // or contains a single element 
        if (n == 0 || n == 1) 
            return n; 
      
        int []temp = new int[n]; 
          
        // Start traversing elements 
        int j = 0; 
          
        for (int i = 0; i < n - 1; i++) 
          
            // If current element is not equal 
            // to next element then store that 
            // current element 
            if (arr[i] != arr[i+1]) 
                temp[j++] = arr[i]; 
          
        // Store the last element as 
        // whether it is unique or  
        // repeated, it hasn't 
        // stored previously 
        temp[j++] = arr[n-1];  
          
        // Modify original array 
        for (int i = 0; i < j; i++) 
            arr[i] = temp[i]; 
      
        return j; 
    } 
      
    public static void Main ()  
    { 
        int []arr = {1, 2, 2, 3, 4, 4, 4, 5, 5}; 
        int n = arr.Length; 
          
        n = removeDuplicates(arr, n); 
  
        // Print updated array 
        for (int i = 0; i < n; i++) 
            Console.Write(arr[i] + " "); 
    } 
} 
  
// This code is contributed by nitin mittal. 

Output:

1 2 3 4 5

Time Complexity : O(n)
Auxiliary Space : O(n)

Method 2: (Constant extra space)
Just maintain a separate index for same array as maintained for different array in Method 1.

C++

// C++ program to remove duplicates in-place 
#include<iostream> 
using namespace std; 
  
// Function to remove duplicate elements 
// This function returns new size of modified 
// array. 
int removeDuplicates(int arr[], int n) 
{ 
    if (n==0 || n==1) 
        return n; 
  
    // To store index of next unique element 
    int j = 0; 
  
    // Doing same as done in Method 1 
    // Just maintaining another updated index i.e. j 
    for (int i=0; i < n-1; i++) 
        if (arr[i] != arr[i+1]) 
            arr[j++] = arr[i]; 
  
    arr[j++] = arr[n-1]; 
  
    return j; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    // removeDuplicates() returns new size of 
    // array. 
    n = removeDuplicates(arr, n); 
  
    // Print updated array 
    for (int i=0; i<n; i++) 
        cout << arr[i] << " "; 
  
    return 0; 
} 

Java

// simple java program to remove 
// duplicates 
  
class Main 
{ 
    // Function to remove duplicate elements 
    // This function returns new size of modified 
    // array. 
    static int removeDuplicates(int arr[], int n) 
    { 
        if (n == 0 || n == 1) 
            return n; 
       
        // To store index of next unique element 
        int j = 0; 
       
        // Doing same as done in Method 1 
        // Just maintaining another updated index i.e. j 
        for (int i = 0; i < n-1; i++) 
            if (arr[i] != arr[i+1]) 
                arr[j++] = arr[i]; 
       
        arr[j++] = arr[n-1]; 
       
        return j; 
    } 
      
    public static void main (String[] args)  
    { 
        int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5}; 
        int n = arr.length; 
          
        n = removeDuplicates(arr, n); 
   
        // Print updated array 
        for (int i=0; i<n; i++) 
           System.out.print(arr[i]+" "); 
    } 
} 
  
/* This code is contributed by Harsh Agarwal */

Python3

# Python3 program to remove 
# duplicate elements 
  
# This function returns new  
# size of modified array 
def removeDuplicates(arr, n): 
    if n == 0 or n == 1: 
        return n 
  
    # To store index of next 
    # unique element 
    j = 0
  
    # Doing same as done 
    # in Method 1 Just 
    # maintaining another  
    # updated index i.e. j 
    for i in range(0, n-1): 
        if arr[i] != arr[i+1]: 
            arr[j] = arr[i] 
            j += 1
  
    arr[j] = arr[n-1] 
    j += 1
    return j 
  
# Driver code 
arr = [1, 2, 2, 3, 4, 4, 4, 5, 5] 
n = len(arr) 
  
# removeDuplicates() returns 
# new size of array. 
n = removeDuplicates(arr, n) 
  
# Print updated array 
for i in range(0, n): 
    print (" %d "%(arr[i]), end = " ") 

C#

// simple C# program to remove 
// duplicates 
using System; 
  
class GfG { 
      
    // Function to remove duplicate 
    // elements This function returns  
    // new size of modified array. 
    static int removeDuplicates(int []arr, int n) 
    { 
          
        if (n == 0 || n == 1) 
            return n; 
      
        // To store index of next 
        // unique element 
        int j = 0; 
      
        // Doing same as done in Method 1 
        // Just maintaining another updated 
        // index i.e. j 
        for (int i = 0; i < n - 1; i++) 
            if (arr[i] != arr[i + 1]) 
                arr[j++] = arr[i]; 
      
        arr[j++] = arr[n - 1]; 
      
        return j; 
    } 
      
    public static void Main ()  
    { 
        int []arr = {1, 2, 2, 3, 4, 4, 
                                 4, 5, 5}; 
        int n = arr.Length; 
          
        n = removeDuplicates(arr, n); 
  
        // Print updated array 
        for (int i = 0; i < n; i++) 
            Console.Write(arr[i] + " "); 
    } 
} 
  
// This code is contributed by parashar. 

Output:

1 2 3 4 5

Time Complexity : O(n)
Auxiliary Space : O(1)

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

C++

Java

Python3

C#

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