Reduce the string by removing K consecutive identical characters

// 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 = "w3wiki";
    int k = 2;
    Solution obj;
    cout << obj.remove_k_char(k, s) << "\n";

    return 0;
}

// This Code has been contributed by shubhamm050402

#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 = "w3wiki";
    string ans = reduced_String(k, st);
    cout << ans << endl;
    return 0;
}

#include <iostream>
#include <string>

using namespace std;

string solve(string s, int k) {
    string ans = "";
    for (int i = 0; i < s.size(); i++) {
        int j = i;
        string temp = "";
        while (j < s.size() && s[i] == s[j]) {
            temp += s[i];
            j++;
        }
        if (j - i != k)
            ans += temp;
        i = j - 1;
    }
    return ans;
}

int main() {
    int k;
    string str;

    // Input
    cout << "Enter the string: ";
    cin >> str;
    cout << "Enter the value of K: ";
    cin >> k;

    // Function call
    string reducedString = solve(str, k);

    // Output
    cout << "Reduced string: " << reducedString << endl;

    return 0;
}