Recursively Remove all Adjacent Duplicate Characters in C++

Remove all Adjacent Duplicate Characters Recursively in C++

Here, in this page we will discuss the program to remove all adjacent duplicate characters recursively in C++ programming language. We are given with an string and need to return the string after removing all its adjacent duplicate characters.

Example :

  • Input : str = “aabaabccded”
  • Output : e
Remove all Adjacent Duplicate Characters recursively in C++

Method 1 :

In this algorithm we will discuss the recursive way to remove all  duplicate character from the given input string s.

  • First we check whether the String remStr has the repeated character that matches the last char of the parent String.
  • If that is so then we have to keep removing that character before concatenating string s and string remStr.
Remove all adjacent duplicates

Code to Recursively Remove all Adjacent Duplicate in C++

Run
#include <bits/stdc++.h>
using namespace std;

string removeDuplicates(string s, char ch)
{

   // Base condition
   if (s.length() <= 1) {
       return s;
   }

   int i = 0;
   while (i < s.length()) {
      if (i + 1 < s.length() && s[i] == s[i + 1]) {
         int j = i;
         while (j + 1 < s.length() && s[j] == s[j + 1]) { 
             j++; 
             
         } 
         char lastChar = i > 0 ? s[i - 1] : ch;

         string remStr = removeDuplicates(
         s.substr(j + 1, s.length()), lastChar);

         s = s.substr(0, i);
         int k = s.length(), l = 0;

         while (remStr.length() > 0 && s.length() > 0 && remStr[0] == s[s.length() - 1]) 
         {

             while (remStr.length() > 0 && remStr[0] != ch && remStr[0] == s[s.length() - 1]) {
                   remStr = remStr.substr(1, remStr.length());
             }
             s = s.substr(0, s.length() - 1);
         }
         s = s + remStr;
         i = j;
    }
    else {
         i++;
    }
  }
  return s;
}

// Driver Code
int main()
{

   string str1 = "aabbaaacdffd";
   cout << removeDuplicates(str1, ' ') << endl;
}
Output :

c

Method 2 (Using Stack) :

In this method we will use Stack as it follows last in first out (LIFO) mechanism.

  • Declare a stack of char type say stck.
  • Now, iterate the string str and for every i-th character,
  • Check if stck.empty() == true and stck.top() != str[i]. Then, push str[i] into stack.
  • Otherwise pop() the element from the stack.
  • Now, check if stck.empty() == true then print(“Empty String”)
  • Otherwise, declare a string say result=””
  • Run a loop till stck.empty()==false,
  • And set, result = stck.top()+ result.
  • At last print(result)

Code in C++

Run
#include<bits/stdc++.h>
using namespace std;

//Function to remove Duplicates
string removeDuplicates(string str){

   stack<char>stck;

   for(int i=0; i< str.length(); i++){

      if(stck.empty() or stck.top()!= str[i])
        stck.push(str[i]);

      else stck.pop();
   }

   if(stck.empty())
     return "Empty String";

    string result = "";

    while(!stck.empty()){
         result = stck.top()+result;
         stck.pop();
    }

    return result;
}


//Driver Code
int main(){
    string str="aacbbcaaded";
    cout<< removeDuplicates(str);
}

Output :

ded

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One comment on “Recursively Remove all Adjacent Duplicate Characters in C++”


  • abhayagrwal2002

    // when you try to run string = “acaabbbceddd” this code output is wrong is wrong.
    // Maybe the correct code is this :-

    #include
    using namespace std;

    //Function to remove Duplicates
    string removeDuplicates(string str){

    stackstck;

    for(int i=0; i< str.length(); i++){

    if(stck.empty() or stck.top()!= str[i])
    stck.push(str[i]);

    else {
    while(str[i]==stck.top()){
    i++;
    }
    i–;
    stck.pop();
    }
    }

    if(stck.empty())
    return "Empty String";

    string result = "";

    while(!stck.empty()){
    result = stck.top()+result;
    stck.pop();
    }

    return result;
    }

    //Driver Code
    int main(){
    string str="acaabbbceddd";
    cout<< removeDuplicates(str);
    }