Repeated Deletion Sequel - Amazon Top Interview Questions

Problem Statement :

Given a string s and an integer k, repeatedly delete the earliest k consecutive duplicate characters.


k, n ≤ 100,000 where n is the length of s.

Example 1


s = "baaabbdddd"

k = 3




First we delete the "a"s to get "bbbdddd". Then we delete the "b"s to get "dddd". Then we delete three of 
the four "d"s to get "d"

Solution :


                        Solution in C++ :

string solve(string s, int k) {
    if (k == 1) return "";
    vector<pair<int, char>> stack{{0, '#'}};

    for (auto &c : s) {
        if (stack.back().second != c) {
            stack.push_back({1, c});
        } else if (++stack.back().first == k) {

    string result = "";

    for (int i = 1; i < stack.size(); i++) {
        for (int j = 0; j < stack[i].first; j++) {
            result += stack[i].second;

    return result;

                        Solution in Java :

import java.util.*;
class pair { // created  a class pair to store char and its count.
    char val;
    int num;
    pair(char val, int num) {
        this.val = val;
        this.num = num;
class Solution {
    public String solve(String s, int k) {
        Stack<pair> st = new Stack<>(); // stack which has pair as attribute.

        for (int i = 0; i < s.length(); i++) {
            if (st.isEmpty()) { // check if stack is empty then simply push with count 1.
                st.push(new pair(s.charAt(i), 1));
            } else if (s.charAt(i)
                == st.peek().val) { // check if incoming value is equal to peak value or not.
                st.push(new pair(s.charAt(i), st.peek().num + 1));
                // if its equal to stack top value, we'll insert this into stack with count 1+ of
                // peek.
            } else {
                st.push(new pair(
                    s.charAt(i), 1)); // last entered char was different so simply push with 1.
            if (st.peek().num == k) { // if we arrive to that point where top value count = k.
                int x = st.peek().num;
                while (x > 0) { // we will pop all k values from stack.
        String fin = ""; // answer string.
        while (st.size() > 0) { // while we dont get empty stack.
            fin = st.pop().val + fin;
        return fin;

                        Solution in Python : 
class Solution:
    def solve(self, s, k):
        stack = []

        for c in s:
            if stack and c == stack[-1][0]:
                stack[-1][1] += 1
                stack.append([c, 1])

            if stack[-1][1] == k:

        return "".join([x * f for x, f in stack])

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