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Reduce the string by removing K consecutive identical characters

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Given a string str and an integer K, the task is to reduce the string by applying the following operation any number of times until it is no longer possible:

Choose a group of K consecutive identical characters and remove them from the string.

Finally, print the reduced string.

Examples:  

Input: K = 2, str = “geeksforgeeks” 
Output: gksforgks 
Explanation: After removal of both occurrences of the substring “ee”, the string reduces to “gksforgks”.

Input: K = 3, str = “qddxxxd” 
Output:
Explanation: 
Removal of “xxx” modifies the string to “qddd”. 
Again, removal of “ddd”modifies the string to “q”. 

Recommended Practice

Approach: This problem can be solved using the Stack data structure. Follow the steps below to solve the problem:

Below is the implementation of the above approach: 

C++

// CPP program for the above approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Basic Approach is to create a Stack that store the
// Character and its continuous repetition number This is
// done using pair<char,int> Further we check at each
// iteration, whether the character matches the top of stack
// if it does then check for number of repetitions
// else add to top of stack with count 1
 
class Solution {
public:
    string remove_k_char(int k, string s)
    {
 
        // Base Case
        // If k=1 then all characters
        // can be removed at each
        // instance
        if (k == 1)
            return "";
        // initialize string
        string output = "";
 
        // create a stack using pair<> for storing each
        // character and corresponding
        // repetition
        stack<pair<char, int> > stk;
 
        // iterate through the string
        for (int i = 0; i < s.length(); i++) {
 
            // if stack is empty then simply add the
            // character with count 1 else check if
            // character is same as top of stack
            if (stk.empty() == true) {
                stk.push(make_pair(s[i], 1));
            }
            else {
 
                // if character at top of stack is same as
                // current character increase the number of
                // repetitions in the top of stack by 1
                if (s[i] == (stk.top()).first) {
                    pair<char, int> P = stk.top();
                    stk.pop();
                    P.second++;
                    if (P.second == k)
                        continue;
                    else
                        stk.push(P);
                }
                else {
 
                    // if character at top of stack is not
                    // same as current character push the
                    // character along with count 1 into the
                    // top of stack
                    stk.push(make_pair(s[i], 1));
                }
            }
        }
 
        // Iterate through the stack
        // Use string(int,char) in order to replicate the
        // character multiple times and convert into string
        // then add in front of output string
        while (!stk.empty()) {
            if (stk.top().second > 1) {
                // if Frequency of current character greater
                // than 1(let m),then append that character
                // m times in output string
                int count = stk.top().second;
                while (count--)
                    output += stk.top().first;
            }
            else {
                output += stk.top().first;
            }
            stk.pop();
        }
        reverse(output.begin(), output.end());
        return output;
    }
};
 
// Driver Code
int main()
{
 
    string s = "geeksforgeeks";
    int k = 2;
    Solution obj;
    cout << obj.remove_k_char(k, s) << "\n";
 
    return 0;
}
 
// This Code has been contributed by shubhamm050402

                    

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
typedef struct {
    char ch;
    int freq;
} Pair;
 
Pair createPair(char ch, int freq)
{
    Pair p;
    p.ch = ch;
    p.freq = freq;
    return p;
}
 
void push(Pair* stack, int* top, char ch, int freq)
{
    Pair p = createPair(ch, freq);
    stack[++(*top)] = p;
}
 
Pair pop(Pair* stack, int* top) { return stack[(*top)--]; }
 
char* remove_k_char(int k, char* s)
{
    if (k == 1)
        return "";
 
    Pair stack[strlen(s)];
    int top = -1;
 
    for (int i = 0; i < strlen(s); i++) {
        if (top == -1 || stack[top].ch != s[i])
            push(stack, &top, s[i], 1);
        else {
            stack[top].freq++;
            if (stack[top].freq == k)
                pop(stack, &top);
        }
    }
 
    int resultLength = top + 1;
    char* result
        = (char*)malloc((resultLength + 1) * sizeof(char));
 
    for (int i = 0; i < resultLength; i++)
        result[i] = stack[i].ch;
 
    result[resultLength] = '\0';
    return result;
}
 
int main()
{
    char s[] = "geeksforgeeks";
    int k = 2;
 
    char* result = remove_k_char(k, s);
    printf("%s\n", result);
 
    free(result);
 
    return 0;
}

                    

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to find the reduced string
    public static String reduced_String(int k, String s)
    {
 
        // Base Case
        if (k == 1) {
            // all elements remove,send empty string
            return "";
        }
 
        // Creating a stack of type Pair
        Stack<Pair> st = new Stack<Pair>();
 
        // Length of the string S
        int l = s.length();
        int ctr = 0;
 
        // iterate through the string
        for (int i = 0; i < l; i++) {
            // if stack is empty then simply add the
            // character with count 1 else check if
            // character is same as top of stack
            if (st.size() == 0) {
                st.push(new Pair(s.charAt(i), 1));
                continue;
            }
 
            // if character at top of stack is same as
            // current character increase the number of
            // repetitions in the top of stack by 1
            if (st.peek().c == s.charAt(i)) {
                Pair p = st.peek();
                st.pop();
                p.ctr += 1;
                if (p.ctr == k) {
                    continue;
                }
                else {
                    st.push(p);
                }
            }
            else {
                st.push(new Pair(s.charAt(i), 1));
            }
        }
 
        // iterate through the stack
        // append characters in String
 
        StringBuilder output = new StringBuilder();
 
        while (st.size() > 0) {
            char c = st.peek().c;
            int cnt = st.peek().ctr;
            // If frequency of a character is cnt, then
            // append that character to cnt times in String
            while (cnt-- > 0)
                output.append(String.valueOf(c));
            st.pop();
        }
        output.reverse();
        return output.toString();
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int k = 2;
        String st = "geeksforgeeks";
        String ans = reduced_String(k, st);
        System.out.println(ans);
    }
 
    // Pair Class
    static class Pair {
        char c;
        int ctr;
        Pair(char c, int ctr)
        {
            this.c = c;
            this.ctr = ctr;
        }
    }
}
 
// This Code has been contributed by shubhamm050402

                    

Python3

# Python3 implementation of the approach
 
# Pair class to store character and freq
class Pair:
    def __init__(self,c ,ctr):
        self.c= c
        self.ctr = ctr
 
class Solution:
     
    # Function to find the reduced string
    def reduced_String(self , k , s):
         
        #Base Case
        if (k == 1):
            return ""
 
        # Creating a stack of type Pair
        st = []
     
        # iterate through given string
        for i in range(len(s)):
             
            # if stack is empty then simply add the
            # character with count 1 else check if
            # character is same as top of stack
            if (len(st) == 0):
                st.append((Pair(s[i] , 1)))
                continue
                 
             
            # if character at top of stack is same as
            # current character increase the number of
            # repetitions in the top of stack by 1
            if (st[-1].c == s[i]):
                 
                pair = st.pop()
                pair.ctr +=1
                 
                if (pair.ctr == k):
                    continue
                 
                else:
                    st.append(pair)
     
             
            else:
                 
                # if character at top of stack is not
                # same as current character push the
                # character along with count 1 into the
                # top of stack
                st.append((Pair(s[i] , 1)))
     
     
        # Iterate through the stack
        # Use string(int,char) in order to replicate the
        # character multiple times and convert into string
        # then add in front of output string
        ans = ""
        while(len(st) > 0):
             
            c = st[-1].c
            cnt = st[-1].ctr
             
            while(cnt >0):
                ans  = c + ans
                cnt -= 1
             
            st.pop()
         
        return (ans)
 
# Driver code
if __name__ == "__main__":
     
    k = 2
    s = "geeksforgeeks"
    obj = Solution()
    print(obj.reduced_String(k,s))
 
    # This code is contributed by chantya17.

                    

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;
public class GFG {
 
    // Function to find the reduced string
    public static String reduced_String(int k, String s)
    {
        // Base Case
        if (k == 1) {
             
            return "";
        }
 
        // Creating a stack of type Pair
        Stack<Pair> st = new Stack<Pair>();
 
        // Length of the string S
        int l = s.Length;
      
        // iterate through the string
        for (int i = 0; i < l; i++) {
 
            // if stack is empty then simply add the
            // character with count 1 else check if
            // character is same as top of stack
            if (st.Count == 0) {
                st.Push(new Pair(s[i], 1));
                continue;
            }
 
            // if character at top of stack is same as
            // current character increase the number of
            // repetitions in the top of stack by 1
            if (st.Peek().c == s[i]) {
                Pair p = st.Peek();
                st.Pop();
                p.ctr += 1;
                if (p.ctr == k) {
                    continue;
                }
                else {
                    st.Push(p);
                }
            }
            else {
 
                // if character at top of stack is not
                // same as current character push the
                // character along with count 1 into the
                // top of stack
                st.Push(new Pair(s[i], 1));
            }
        }
 
        // iterate through the stack
        // Use string(int,char) in order to replicate the
        // character multiple times and convert into string
        // then add in front of output string
        String ans = "";
        while (st.Count > 0) {
            char c = st.Peek().c;
            int cnt = st.Peek().ctr;
            while (cnt-- > 0)
                ans = c + ans;
            st.Pop();
        }
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int k = 2;
        String st = "geeksforgeeks";
        String ans = reduced_String(k, st);
        Console.Write(ans);
    }
 
    public class Pair {
        public char c;
        public int ctr;
        public Pair(char c, int ctr)
        {
            this.c = c;
            this.ctr = ctr;
        }
    }
}
// This code has been contributed by 29AjayKumar

                    

Javascript

<script>
// Javascript implementation of the approach
 
class Pair
{
    constructor(c,ctr)
    {
        this.c = c;
            this.ctr = ctr;
    }
}
 
// Function to find the reduced string
function reduced_String(k,s)
{
    // Base Case
        if (k == 1) {
            let ans = "";
            return ans;
        }
  
        // Creating a stack of type Pair
        let st = [];
  
        // Length of the string S
        let l = s.length;
        let ctr = 0;
  
        // iterate through the string
        for (let i = 0; i < l; i++) {
  
            // if stack is empty then simply add the
            // character with count 1 else check if
            // character is same as top of stack
            if (st.length == 0) {
                st.push(new Pair(s[i], 1));
                continue;
            }
  
            // if character at top of stack is same as
            // current character increase the number of
            // repetitions in the top of stack by 1
            if (st[st.length-1].c == s[i]) {
                let p = st[st.length-1];
                st.pop();
                p.ctr += 1;
                if (p.ctr == k) {
                    continue;
                }
                else {
                    st.push(p);
                }
            }
            else {
  
                // if character at top of stack is not
                // same as current character push the
                // character along with count 1 into the
                // top of stack
                st.push(new Pair(s[i], 1));
            }
        }
  
        // iterate through the stack
        // Use string(int,char) in order to replicate the
        // character multiple times and convert into string
        // then add in front of output string
        let ans = "";
        while (st.length > 0) {
            let c = st[st.length-1].c;
            let cnt = st[st.length-1].ctr;
            while (cnt-- > 0)
                ans = c + ans;
            st.pop();
        }
        return ans;
}
 
// Driver code
let k = 2;
let st = "geeksforgeeks";
let ans = reduced_String(k, st);
document.write(ans+"<br>");
 
// This code is contributed by rag2127
</script>

                    

Output
gksforgks

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

ANOTHER METHOD:

APPROACH:

  1. First we declare a Stack<Character> to store each character of the string.
  2. Then we iterate over the string.
  3. While iterating we keep a counter variable and keep pushing the character in the stack and poping simultaneously until we get a counter equals k, that implies we have got the sequence of character to remove from the string.
  4. At last we declare a String Builder to concatenate the characters from the stack.

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
 
string reduced_String(int k, string s) {
    stack<char> stk;
    int i = 0;
    while (i < s.length()) {
        char ch = s[i++];
        stk.push(ch);
        int count = 0;
        while (!stk.empty() && stk.top() == ch) {
            count++;
            stk.pop();
        }
        if (count == k)
            continue;
        else {
            while (count > 0) {
                stk.push(ch);
                count--;
            }
        }
    }
    string result = "";
    while (!stk.empty()) {
        result = stk.top() + result;
        stk.pop();
    }
    return result;
}
 
int main() {
    int k = 2;
    string st = "geeksforgeeks";
    string ans = reduced_String(k, st);
    cout << ans << endl;
    return 0;
}

                    

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
#define MAX_SIZE 100
 
// Function to remove k consecutive characters from the
// string
char* reduced_String(int k, char* s)
{
    char stk[MAX_SIZE]; // Stack to store characters
    int top = -1; // Top index of stack
    int i = 0;
 
    while (s[i] != '\0') {
        char ch = s[i++];
        stk[++top] = ch;
 
        int count = 0;
        while (top >= k - 1 && stk[top] == ch) {
            count++;
            top--;
        }
 
        if (count == k)
            continue;
        else {
            while (count > 0) {
                stk[++top] = ch;
                count--;
            }
        }
    }
 
    char* result = (char*)malloc(
        (top + 2)
        * sizeof(char)); // Allocate memory for the result
 
    int resultIndex = 0;
    while (top >= 0) {
        result[resultIndex++] = stk[top];
        top--;
    }
    result[resultIndex]
        = '\0'; // Null-terminate the result string
 
    return result;
}
 
// Driver code
int main()
{
    int k = 2;
    char st[] = "skgrofskg";
 
    char* ans = reduced_String(k, st);
    printf("%s\n", ans);
 
    free(ans); // Free the memory allocated for the result
 
    return 0;
}

                    

Java

import java.io.*;
import java.util.*;
 
class GFG {
  public static String reduced_String(int k, String s)
  {
      // Your code goes here
      Stack<Character> stk = new Stack<Character>();
      int i = 0;
      while (i < s.length()) {
          char ch = s.charAt(i++);
          stk.push(ch);
          int count = 0;
          while ((stk.size() > 0) && (stk.peek() == ch)) {
              count++;
              stk.pop();
          }
          if (count == k)
              continue;
          else {
              while (count > 0) {
                  stk.push(ch);
                  count--;
              }
          }
      }
      StringBuilder sb = new StringBuilder();
      for (char ch : stk)
          sb = sb.append(ch);
      return sb.toString();
  }
      public static void main(String[] args)
      {
          int k = 2;
          String st = "geeksforgeeks";
          String ans = reduced_String(k, st);
          System.out.println(ans);
      }
  }
//This code is contributed by Raunak Singh

                    

C#

// Importing required libraries
using System;
using System.Collections.Generic;
 
class Program {
 
    // Function to reduce the string
    static string ReducedString(int k, string s)
    {
 
        // Creating an empty stack
        // to hold characters
        Stack<char> stk = new Stack<char>();
        int i = 0;
 
        // Loop through each character
        // of the input string
        while (i < s.Length) {
            char ch = s[i++];
            stk.Push(ch);
 
            int count = 0;
 
            // Count the number of consecutive
            // occurrences of  current character
            while (stk.Count > 0 && stk.Peek() == ch) {
                count++;
                stk.Pop();
            }
 
            // If the count is equal to k,
            // skip to the next character
            if (count == k)
                continue;
            else {
 
                // Otherwise, push back the
                // characters that were popped
                while (count > 0) {
                    stk.Push(ch);
                    count--;
                }
            }
        }
 
        // Build the final reduced string
        // by popping all characters
        // from the stack
        string result = "";
        while (stk.Count > 0) {
            result = stk.Pop() + result;
        }
 
        return result;
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        int k = 2;
        string st = "geeksforgeeks";
 
        string ans = ReducedString(k, st);
 
        Console.WriteLine(ans);
    }
}

                    

Javascript

// Function to reduce the string
function reducedString(k, s) {
// Creating an empty stack to hold characters
  const stk = [];
  let i = 0;
 
// Loop through each character of the input string
   while (i < s.length) {
   const ch = s[i++];
   stk.push(ch);
 
   let count = 0;
 
// Count the number of consecutive occurrences of current character
while (stk.length > 0 && stk[stk.length - 1] === ch) {
  count++;
  stk.pop();
}
 
// If the count is equal to k, skip to the next character
if (count === k) {
  continue;
} else {
  // Otherwise, push back the characters that were popped
  while (count > 0) {
    stk.push(ch);
    count--;
  }
}
}
 
// Build the final reduced string by popping all characters from the stack
 let result = "";
 while (stk.length > 0) {
 result = stk.pop() + result;
}
 
return result;
}
 
// Driver Code
const k = 2;
const st = "geeksforgeeks";
const ans = reducedString(k, st);
console.log(ans);

                    

Output
gksforgks

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


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Last Updated : 09 May, 2023
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