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Remove all consecutive duplicates from the string

  • Difficulty Level : Easy
  • Last Updated : 10 Sep, 2021
Geek Week

Given a string S, remove all the consecutive duplicates. Note that this problem is different from Recursively remove all adjacent duplicates. Here we keep one character and remove all subsequent same characters.

Examples: 

Input  : aaaaabbbbbb
Output : ab

Input : geeksforgeeks
Output : geksforgeks

Input : aabccba
Output : abcba

Recursive Solution:

The above problem can be solved using recursion.  

  1. If the string is empty, return.
  2. Else compare the adjacent characters of the string. If they are same then shift the characters one by one to the left. Call recursion on string S
  3. If they not same then call recursion from S+1 string.

The recursion tree for the string S = aabcca is shown below.  



        aabcca   S = aabcca
        /
       abcca     S = abcca        
       /
      bcca       S = abcca
      /
     cca         S = abcca
     /
    ca           S = abca
   /
  a              S = abca (Output String)
 /
empty string

Below is the implementation of the above approach:  

C++




// Recursive Program to remove consecutive
// duplicates from string S.
#include <bits/stdc++.h>
using namespace std;
 
// A recursive function that removes
// consecutive duplicates from string S
void removeDuplicates(char* S)
{
    // When string is empty, return
    if (S[0] == '\0')
        return;
 
    // if the adjacent characters are same
    if (S[0] == S[1]) {
         
        // Shift character by one to left
        int i = 0;
        while (S[i] != '\0') {
            S[i] = S[i + 1];
            i++;
        }
 
        // Check on Updated String S
        removeDuplicates(S);
    }
 
    // If the adjacent characters are not same
    // Check from S+1 string address
    removeDuplicates(S + 1);
}
 
// Driver Program
int main()
{
    char S1[] = "geeksforgeeks";
    removeDuplicates(S1);
    cout << S1 << endl;
 
    char S2[] = "aabcca";
    removeDuplicates(S2);
    cout << S2 << endl;
 
    return 0;
}

Java




/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    public static String removeConsecutiveDuplicates(String input) {
        if(input.length()<=1)
            return input;
        if(input.charAt(0)==input.charAt(1))
            return removeConsecutiveDuplicates(input.substring(1));
        else
            return input.charAt(0) + removeConsecutiveDuplicates(input.substring(1));
    }
    public static void main(String[] args)
    {
        String S1 = "geeksforgeeks";
        System.out.println(removeConsecutiveDuplicates(S1));
   
        String S2 = "aabcca";
        System.out.println(removeConsecutiveDuplicates(S2));
    }
}

C#




/*package whatever //do not write package name here */
using System;
 
class GFG {
    public static string removeConsecutiveDuplicates(string input) {
        if(input.Length <= 1)
            return input;
        if(input[0] == input[1])
            return removeConsecutiveDuplicates(input.Substring(1));
        else
            return input[0] + removeConsecutiveDuplicates(input.Substring(1));
    }
    public static void Main(String[] args)
    {
        string S1 = "geeksforgeeks";
        Console.WriteLine(removeConsecutiveDuplicates(S1));
   
        string S2 = "aabcca";
        Console.Write(removeConsecutiveDuplicates(S2));
    }
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
 
function removeConsecutiveDuplicates(input)
{
    if(input.length<=1)
            return input;
        if(input[0]==input[1])
            return removeConsecutiveDuplicates(input.substring(1));
        else
            return input[0] +
            removeConsecutiveDuplicates(input.substring(1));
}
 
let S1 = "geeksforgeeks";
document.write(removeConsecutiveDuplicates(S1)+"<br>");
 
let  S2 = "aabcca";
document.write(removeConsecutiveDuplicates(S2)+"<br>");
 
 
// This code is contributed by rag2127
 
</script>
Output
geksforgeks
abca

The worst case time complexity of the above solution is O(n2). The worst case happens when all characters are same. 
 

Iterative Solution :

The idea is to check if current character is equal to the next character or not . If current character is equal to the next character we’ll ignore it and when it is not equal we will add it to our answer. Since, the last element will not checked we will push it at the end of the string. For eg: s=”aaaaa”

when we run the loop str=”” so at the end we’ll add ‘a’ because it is the last element.

C++




// C++ program to remove consecutive
// duplicates from a given string.
#include <bits/stdc++.h>
using namespace std;
 
// A iterative function that removes 
// consecutive duplicates from string S
string removeDuplicates(string s){
     
   int n = s.length();
   string str="";
   // We don't need to do anything for
   // empty string.
   if (n == 0)
     return str;
 
   // Traversing string
    for(int i=0;i<n-1;i++){
      //checking if s[i] is not same as s[i+1] then add it into str
        if(s[i]!=s[i+1]){
            str+=s[i];
        }
    }
  //Since the last character will not be inserted in the loop we add it at the end
   str.push_back(s[n-1]);
   return str;   
}
  
// Driver Program
int main() {
      
    string s1 = "geeksforgeeks";
    cout << removeDuplicates(s1) << endl;
     
    string s2 = "aabcca";
    cout << removeDuplicates(s2) << endl;
      
    return 0;
}

Java




// Java program to remove consecutive
// duplicates from a given string.
import java.util.*;
 
class GFG
{
 
    // A iterative function that removes
    // consecutive duplicates from string S
    static void removeDuplicates(char[] S)
    {
        int n = S.length;
 
        // We don't need to do anything for
        // empty or single character string.
        if (n < 2)
        {
            return;
        }
 
        // j is used to store index is result
        // string (or index of current distinct
        // character)
        int j = 0;
 
        // Traversing string
        for (int i = 1; i < n; i++)
        {
            // If current character S[i]
            // is different from S[j]
            if (S[j] != S[i])
            {
                j++;
                S[j] = S[i];
            }
        }
        System.out.println(Arrays.copyOfRange(S, 0, j + 1));
    }
 
    // Driver code
    public static void main(String[] args)
    {
        char S1[] = "geeksforgeeks".toCharArray();
        removeDuplicates(S1);
 
        char S2[] = "aabcca".toCharArray();
        removeDuplicates(S2);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 program to remove consecutive
# duplicates from a given string.
 
# A iterative function that removes
# consecutive duplicates from string S
def removeDuplicates(S):
         
    n = len(S)
     
    # We don't need to do anything for
    # empty or single character string.
    if (n < 2) :
        return
         
    # j is used to store index is result
    # string (or index of current distinct
    # character)
    j = 0
     
    # Traversing string
    for i in range(n):
         
        # If current character S[i]
        # is different from S[j]
        if (S[j] != S[i]):
            j += 1
            S[j] = S[i]
     
    # Putting string termination
    # character.
    j += 1
    S = S[:j]
    return S
     
# Driver Code
if __name__ == '__main__':
         
    S1 = "geeksforgeeks"
    S1 = list(S1.rstrip())
    S1 = removeDuplicates(S1)
    print(*S1, sep = "")
         
    S2 = "aabcca"
    S2 = list(S2.rstrip())
    S2 = removeDuplicates(S2)
    print(*S2, sep = "")
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// C# program to remove consecutive
// duplicates from a given string.
using System;
     
class GFG
{
 
    // A iterative function that removes
    // consecutive duplicates from string S
    static void removeDuplicates(char[] S)
    {
        int n = S.Length;
 
        // We don't need to do anything for
        // empty or single character string.
        if (n < 2)
        {
            return;
        }
 
        // j is used to store index is result
        // string (or index of current distinct
        // character)
        int j = 0;
 
        // Traversing string
        for (int i = 1; i < n; i++)
        {
            // If current character S[i]
            // is different from S[j]
            if (S[j] != S[i])
            {
                j++;
                S[j] = S[i];
            }
        }
        char []A = new char[j+1];
        Array.Copy(S,0, A, 0, j + 1);
        Console.WriteLine(A);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        char []S1 = "geeksforgeeks".ToCharArray();
        removeDuplicates(S1);
 
        char []S2 = "aabcca".ToCharArray();
        removeDuplicates(S2);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
    // JavaScript program for the above approach
 
    // duplicates from a given string.
 
    // A iterative function that removes
    // consecutive duplicates from string S
    const removeDuplicates = (s) => {
 
        let n = s.length;
        let str = "";
        // We don't need to do anything for
        // empty string.
        if (n == 0)
            return str;
 
        // Traversing string
        for (let i = 0; i < n - 1; i++) {
            //checking if s[i] is not same as s[i+1] then add it into str
            if (s[i] != s[i + 1]) {
                str += s[i];
            }
        }
        //Since the last character will not be inserted in the loop we add it at the end
         
        str += s[n-1];
        return str;
    }
 
    // Driver Program
    let s1 = "geeksforgeeks";
    document.write(`${removeDuplicates(s1)}<br/>`)
    let s2 = "aabcca";
    document.write(removeDuplicates(s2))
 
 
    // This code is contributed by rakeshsahni
 
</script>
Output
geksforgeks
abca

Time Complexity : O(n) 

Auxiliary Space : O(1)

This article is contributed by Ankur Singh. 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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