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])


View More Similar Problems

Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

View Solution →

Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

View Solution →

Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

View Solution →

Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

View Solution →

Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

View Solution →

Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →