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Reduce the string by removing K consecutive identical characters
  • Difficulty Level : Medium
  • Last Updated : 05 Apr, 2021

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”. 

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) {
                    stk.push(
                        { s[i], stk.top().second + 1 });
                    if (stk.top().second == k) {
                        int x = k;
                        while (x) {
                            stk.pop();
                            x--;
                        }
                    }
                }
                else {
 
                    // if character at top of stack is not
                    // same as current charcter 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()) {
            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 Navansh Goel

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) {
            String ans = "";
            return ans;
        }
 
        // 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 {
 
                // if character at top of stack is not
                // same as current charcter push the
                // character along with count 1 into the
                // top of stack
                st.push(new Pair(s.charAt(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.size() > 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);
        System.out.println(ans);
    }
 
    static class Pair {
        char c;
        int ctr;
        Pair(char c, int ctr)
        {
            this.c = c;
            this.ctr = ctr;
        }
    }
}

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 charcter 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
Output
gksforgks

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

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