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.

Constraints

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

Example 1

Input

s = "baaabbdddd"

k = 3

Output

"d"

Explanation

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 :



title-img 




                        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) {
            stack.pop_back();
        }
    }

    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.
                    st.pop();
                    x--;
                }
            }
        }
        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
            else:
                stack.append([c, 1])

            if stack[-1][1] == k:
                stack.pop()

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


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