Remove Duplicates From Sorted Array 334 Easy

Remove duplicates from sorted array

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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

Method 1: (Using extra space)

1. Create an auxiliary array temp[] to store unique elements.
2. 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.
3. Copy j elements from temp[] to arr[] and return j

Implementation:

C++

#include <iostream>

using namespace std;

int removeDuplicates( int arr[], int n)

{

if (n == 0 || n == 1)

return n;

int temp[n];

int j = 0;

for ( int i = 0; i < n - 1; i++)

if (arr[i] != arr[i + 1])

temp[j++] = arr[i];

temp[j++] = arr[n - 1];

for ( int i = 0; i < j; i++)

arr[i] = temp[i];

return j;

}

int main()

{

int arr[] = { 1, 2, 2, 3, 4, 4, 4, 5, 5 };

int n = sizeof (arr) / sizeof (arr[0]);

n = removeDuplicates(arr, n);

for ( int i = 0; i < n; i++)

cout << arr[i] << " " ;

return 0;

}

C

#include <stdio.h>

int removeDuplicates( int arr[], int n)

{

if (n == 0 || n == 1)

return n;

int temp[n];

int j = 0;

for ( int i = 0; i < n - 1; i++)

if (arr[i] != arr[i + 1])

temp[j++] = arr[i];

temp[j++] = arr[n - 1];

for ( int i = 0; i < j; i++)

arr[i] = temp[i];

return j;

}

int main()

{

int arr[] = { 1, 2, 2, 3, 4, 4, 4, 5, 5 };

int n = sizeof (arr) / sizeof (arr[0]);

n = removeDuplicates(arr, n);

for ( int i = 0; i < n; i++)

printf ( "%d " , arr[i]);

return 0;

}

Java

class Main {

static int removeDuplicates( int arr[], int n)

{

if (n == 0 || n == 1 )

return n;

int [] temp = new int [n];

int j = 0 ;

for ( int i = 0 ; i < n - 1 ; i++)

if (arr[i] != arr[i + 1 ])

temp[j++] = arr[i];

temp[j++] = arr[n - 1 ];

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);

for ( int i = 0 ; i < n; i++)

System.out.print(arr[i] + " " );

}

}

Python3

def removeDuplicates(arr, n):

if n = = 0 or n = = 1 :

return n

temp = list ( range (n))

j = 0 ;

for i in range ( 0 , n - 1 ):

if arr[i] ! = arr[i + 1 ]:

temp[j] = arr[i]

j + = 1

temp[j] = arr[n - 1 ]

j + = 1

for i in range ( 0 , j):

arr[i] = temp[i]

return j

arr = [ 1 , 2 , 2 , 3 , 4 , 4 , 4 , 5 , 5 ]

n = len (arr)

n = removeDuplicates(arr, n)

for i in range (n):

print ( "%d" % (arr[i]), end = " " )

C#

using System;

class GFG {

static int removeDuplicates( int []arr, int n)

{

if (n == 0 || n == 1)

return n;

int []temp = new int [n];

int j = 0;

for ( int i = 0; i < n - 1; i++)

if (arr[i] != arr[i+1])

temp[j++] = arr[i];

temp[j++] = arr[n-1];

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);

for ( int i = 0; i < n; i++)

Console.Write(arr[i] + " " );

}

}

Javascript

<script>

function removeDuplicates(arr, n)

{

if (n==0 || n==1)

return n;

var temp = new Array(n);

var j = 0;

for ( var i=0; i<n-1; i++)

if (arr[i] != arr[i+1])

temp[j++] = arr[i];

temp[j++] = arr[n-1];

for ( var i=0; i<j; i++)

arr[i] = temp[i];

return j;

}

var arr = [1, 2, 2, 3, 4, 4, 4, 5, 5];

var n = arr.length;

n = removeDuplicates(arr, n);

for ( var i=0; i<n; i++)

document.write( arr[i]+ " " );

</script>

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.

Implementation:

C++

#include<iostream>

using namespace std;

int removeDuplicates( int arr[], int n)

{

if (n==0 || n==1)

return n;

int j = 0;

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;

}

int main()

{

int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5};

int n = sizeof (arr) / sizeof (arr[0]);

n = removeDuplicates(arr, n);

for ( int i=0; i<n; i++)

cout << arr[i] << " " ;

return 0;

}

Java

class Main

{

static int removeDuplicates( int arr[], int n)

{

if (n == 0 || n == 1 )

return n;

int j = 0 ;

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);

for ( int i= 0 ; i<n; i++)

System.out.print(arr[i]+ " " );

}

}

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#

using System;

class GfG {

static int removeDuplicates( int []arr, int n)

{

if (n == 0 || n == 1)

return n;

int j = 0;

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);

for ( int i = 0; i < n; i++)

Console.Write(arr[i] + " " );

}

}

Javascript

<script>

function removeDuplicates(arr , n) {

if (n == 0 || n == 1)

return n;

var j = 0;

for (i = 0; i < n - 1; i++)

if (arr[i] != arr[i + 1])

arr[j++] = arr[i];

arr[j++] = arr[n - 1];

return j;

}

var arr = [ 1, 2, 2, 3, 4, 4, 4, 5, 5 ];

var n = arr.length;

n = removeDuplicates(arr, n);

for (i = 0; i < n; i++)

document.write(arr[i] + " " );

</script>

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Source: https://www.geeksforgeeks.org/remove-duplicates-sorted-array/

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