Simple Text Editor – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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1 Month Preparation Kit Solutions – HackerRank
C++ Simple Text Editor HackerRank Solution
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <limits>
#include <tuple>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cassert>
using namespace std;
typedef struct query
{
int type, remove;
string add;
}query;
char stak[1000003];
int topp = 0;
void push(string s)
{
for(int i = 0 ; i < s.length(); i++)
{
stak[++topp] = s[i];
}
assert(topp <= 1000000);
}
string pop(int remove)
{
assert(remove >= 0 && remove <= topp);
string popped = "";
int del = remove;
for (int i = topp; del > 0; i--, del--) {
popped = stak[i] + popped;
}
topp -= remove;
return popped;
}
int main()
{
int t, type, remove, k;
stack<query> q_stack;
cin>>t;
while(t--)
{
cin>>type;
if(type == 1)
{
string add;
cin>>add;
push(add);
query last;
last.type = type;
last.add = add;
q_stack.push(last);
}
else if(type == 2)
{
cin>>remove;
string popped = pop(remove);
query last;
last.type = type;
last.remove = remove;
last.add = popped;
q_stack.push(last);
}
else if(type == 3)
{
cin>>k;
assert(k >= 1 && k <= topp);
cout<<stak[k]<<endl;
}
else
{
query last = q_stack.top();
q_stack.pop();
if(last.type == 1)
{
int remove = (last.add).length();
string popped = pop(remove);
}
else
{
push(last.add);
assert(topp >= 1 && topp <= 1000000);
}
}
}
return 0;
}
Java Simple Text Editor HackerRank Solution
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long Q = sc.nextInt();
int tag, k;
String last, newString;
Stack<String> stack = new Stack<>();
while(Q-->0){
tag = sc.nextInt();
switch (tag){
case 1:
//append W
last = stack.size() > 0 ? stack.peek() : "";
newString = last + sc.next();
//System.out.println(tag + " " + newString);
stack.push(newString);
break;
case 2:
//erase last k character of S
k = sc.nextInt();
last = stack.peek();
newString = last.substring(0, last.length()-k);
//System.out.println(tag + " " + newString);
stack.push(newString);
break;
case 3:
//return kth character of S
k = sc.nextInt()-1;
if(stack.size() > 0) {
last = stack.peek();
String c = String.valueOf(last.charAt(k));
//System.out.println(tag + " " + c);
System.out.println(c);
}
break;
case 4:
//undo
//System.out.println(tag + " " + stack.peek());
stack.pop();
break;
}
}
}
}
Python 3 Simple Text Editor HackerRank Solution
no_ops = int(input())
ops = []
s = []
for i in range(no_ops):
one_op = input().split(' ')
ops.append(one_op)
from copy import copy
history = []
for op in ops:
#print(s)
#print(op)
if op[0] == '1':
to_append = op[1]
history.append(copy(s))
s.extend(to_append)
elif op[0] == '2':
k = int(op[1])
history.append(copy(s))
del s[-k:]
elif op[0] == '3':
k = int(op[1])
print(s[k-1])
elif op[0] == '4':
s = history.pop()
JavaScript Simple Text Editor HackerRank Solution
function processData(input) {
var lines = input.split("\n");
var str = "";
var last = [];
for (var i=1; i <= parseInt(lines[0]); i++) {
var command = parseInt(lines[i].split(" ")[0]),
args = lines[i].split(" ")[1];
switch (command) {
case 1:
last.push(str);
str = str + args;
break;
case 2:
last.push(str);
str = str.substring(0, str.length - parseInt(args));
break;
case 3:
console.log(str.charAt(parseInt(args) - 1));
break;
case 4:
str = last.pop();
break;
}
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});
C Simple Text Editor HackerRank Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
int main() {
FILE * in = stdin;
FILE * out = stdout;
int command;
int k;
char buf[1000000];
char ** undos;
int n, i, ptr = 0;
size_t len;
fscanf(in, "%d", &n);
undos = calloc(sizeof(char *), n);
undos[ptr] = malloc(sizeof(char));
strcpy(undos[ptr], "");
for (i = 0; i < n; ++i) {
fscanf(in, "%d", &command);
switch(command) {
case 1:
fscanf(in, "%1000000s\n", buf);
++ptr;
undos[ptr] = malloc(sizeof(char) * (strlen(undos[ptr-1]) + strlen(buf) + 1));
strcpy(undos[ptr], undos[ptr-1]);
strcat(undos[ptr], buf);
break;
case 2:
fscanf(in, "%d", &k);
++ptr;
undos[ptr] = malloc(sizeof(char) * (strlen(undos[ptr - 1]) - k + 1));
len = strlen(undos[ptr - 1]);
memcpy(undos[ptr], undos[ptr-1], strlen(undos[ptr - 1]) - k);
undos[ptr][len - k] = 0;
break;
case 3:
fscanf(in, "%d", &k);
fprintf(out, "%c\n", undos[ptr][k-1]);
break;
case 4:
--ptr;
break;
}
}
return 0;
}
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