Simple Text Editor
Problem Statement :
In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2, reverting S to the state it was in prior to that operation. Input Format The first line contains an integer, Q, denoting the number of operations. Each line i of the Q subsequent lines (where 0 < = i < Q ) defines an operation to be performed. Each operation starts with a single integer, t , denoting a type of operation as defined in the Problem Statement above. If the operation requires an argument, is followed by its space-separated argument. For example, if t = 1 and , W = "abcd" line i will be 1 abcd. Output Format Each operation of type 3 must print the kth character on a new line.
Solution :
Solution in C :
In C++ :
#include <iostream>
#include <string>
#include <stack>
int main() {
std::string text, arg;
int cmd;
std::stack<std::string> history;
std::cin >> cmd;
while (std::cin >> cmd) {
switch (cmd) {
case 1: // Append
std::cin >> arg;
history.push(text);
text.append(arg);
break;
case 2: // Erase
std::cin >> cmd;
history.push(text);
text.erase(text.length() - cmd);
break;
case 3: // Get
std::cin >> cmd;
std::cout << text[cmd - 1] << '\n';
break;
case 4: // Undo
text = std::move(history.top());
history.pop();
break;
}
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = "";
int top = 0;
int q = Integer.parseInt(sc.nextLine());
MyStack stack = new MyStack(q);
for(int i = 0; i < q; ++i){
String st[] = sc.nextLine().split(" ");
int query = Integer.parseInt(st[0]);
if(query == 1){
Node newNode = new Node(query,str.length());
stack.top++;
stack.list[stack.top] = newNode;
str += st[1];
} else if(query == 2){
int k = Integer.parseInt(st[1]);
Node newNode = new Node(query,str.substring(str.length()-k));
stack.top++;
stack.list[stack.top] = newNode;
str = str.substring(0,str.length()-k);
} else if(query == 3){
int index = Integer.parseInt(st[1]);
System.out.println(str.charAt(index-1));
} else if(query == 4){
Node newNode = stack.list[stack.top];
stack.top--;
if(newNode.qtype == 1){
str = str.substring(0,newNode.idx);
} else if(newNode.qtype == 2){
str += newNode.w;
}
}
}
}
}
class MyStack{
Node list[];
int top;
MyStack(int size){
this.list = new Node[size];
this.top = -1;
}
}
class Node{
int qtype;
int idx;
String w;
Node(int x, String y){
this.qtype = x;
this.w = y;
}
Node(int x, int index){
this.qtype = x;
this.idx = index;
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define STACK_SIZE 1000000
#define MAX_W_SIZE ((1000000) + (1))
char* stack[STACK_SIZE];
int sp = -1;
int is_empty() {
return (sp < 0);
}
int is_full() {
return (sp > STACK_SIZE);
}
void push(char* cp) {
if (!is_full()) {
stack[++sp] = cp;
}
}
char* peek() {
char* top = '\0';
if (!is_empty()) {
top = stack[sp];
}
return top;
}
char* pop() {
char* top = peek();
if (top) {
stack[sp--] = '\0';
}
return top;
}
int get_len(char* warg) {
int len = 0;
while(*warg) {
len++; warg++;
}
return len;
}
void do_append(char* warg) {
int len = get_len(warg);
char* current = peek();
if (!current) {
current = (char*) malloc(sizeof(char) * (len + 1));
for (int i = 0; i < len; i++) {
current[i] = warg[i];
}
current[len] = '\0';
push(current);
} else {
int j = 0;
int current_len = get_len(current);
char* current_new = (char*)malloc(sizeof(char) * (current_len + len + 1));
if (current_new) {
for (int i = 0; i < current_len; i++) {
current_new[i] = current[i];
}
for (int i = current_len; i < current_len + len; i++) {
current_new[i] = warg[j++];
}
current_new[current_len + len] = '\0';
push(current_new);
}
}
}
void do_erase(int iarg) {
char* current = peek();
if (current) {
int current_len = get_len(current);
if (current_len >= iarg) {
char* current_new = (char*)malloc(sizeof(char) * (current_len - iarg + 1));
if (current_new) {
for (int i = 0; i < current_len - iarg; i++) {
current_new[i] = current[i];
}
current_new[current_len - iarg] = '\0';
push(current_new);
}
}
}
}
void do_get(int iarg, char* ch) {
char* current = peek();
if (current) {
int current_len = get_len(current);
if (current_len >= iarg) {
*ch = current[iarg - 1];
}
}
}
void do_undo() {
pop();
}
int main() {
int Q;
int op;
int iarg;
char warg[MAX_W_SIZE];
scanf("%d", &Q);
for (int i = 0; i < Q; i++) {
scanf("%d", &op);
if (op == 1) {
scanf("%s", warg);
do_append(warg);
} else if (op == 2) {
scanf("%d", &iarg);
do_erase(iarg);
} else if (op == 3) {
scanf("%d", &iarg);
char ch = '\0';
do_get(iarg, &ch);
fflush(stdout);
printf("%c\n", ch);
} else if (op == 4) {
do_undo();
}
}
return 0;
}
In Python3 :
q = int(input())
stack = ['']
for _ in range(q):
o_type, *par = input().split()
o_type = int(o_type)
if o_type in (1, 2, 3):
par = par[0]
if o_type in (2, 3):
par = int(par)
if o_type == 1:
stack.append(stack[-1] + par)
elif o_type == 2:
stack.append(stack[-1][:-par])
elif o_type == 3:
print(stack[-1][par - 1])
elif o_type == 4:
stack.pop()
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