In this Simple Text Editor HackerRank solution, Implement a simple text editor. The editor initially contains an empty string, . Perform operations of the following types:
- append – Append string to the end of .
- delete – Delete the last characters of .
- print – Print the character of .
- undo – Undo the last (not previously undone) operation of type or , reverting to the state it was in prior to that operation.
Example
operation index S ops[index] explanation ----- ------ ---------- ----------- 0 abcde 1 fg append fg 1 abcdefg 3 6 print the 6th letter - f 2 abcdefg 2 5 delete the last 5 letters 3 ab 4 undo the last operation, index 2 4 abcdefg 3 7 print the 7th characgter - g 5 abcdefg 4 undo the last operation, index 0 6 abcde 3 4 print the 4th character - d
The results should be printed as:
f
g
d
Input Format
The first line contains an integer, , denoting the number of operations.
Each line of the subsequent lines (where ) defines an operation to be performed. Each operation starts with a single integer, (where ), 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 and , line will be 1 abcd.
Constraints
- The sum of the lengths of all in the input .
- The sum of over all delete operations .
- All input characters are lowercase English letters.
- It is guaranteed that the sequence of operations given as input is possible to perform.
Output Format
Each operation of type must print the character on a new line.
Sample Input
STDIN Function ----- -------- 8 Q = 8 1 abc ops[0] = '1 abc' 3 3 ops[1] = '3 3' 2 3 ... 1 xy 3 2 4 4 3 1
Sample Output
c
y
a
Explanation
Initially, is empty. The following sequence of operations are described below:
- . We append to , so .
- Print the character on a new line. Currently, the character is
c. - Delete the last characters in (), so .
- Append to , so .
- Print the character on a new line. Currently, the character is
y. - Undo the last update to , making empty again (i.e., ).
- Undo the next to last update to (the deletion of the last characters), making .
- Print the character on a new line. Currently, the character is
a.
Simple Text Editor HackerRank solution
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in Java
public class Operation
{
public int type;
public string argument;
public Operation(int t, string arg)
{
type = t;
argument = arg;
}
}
static void Main(String[] args) {
int t = Convert.ToInt32(Console.ReadLine().Trim());
List<Operation> ops = new List<Operation>();
for (int tItr = 0; tItr < t; tItr++)
{
string s = Console.ReadLine().TrimEnd();
int type = s[0] - '0';
string arg = string.Empty;
if(type != 4)
{
arg = s.Substring(2, s.Length - 2);
}
Operation oper = new Operation(type, arg);
ops.Add(oper);
}
StringBuilder sb = new StringBuilder();
Stack<Operation> operations = new Stack<Operation>();
for(int i = 0; i < ops.Count; i++)
{
if(ops[i].type == 1)
{
sb.Append(ops[i].argument);
operations.Push(new Operation(5, ops[i].argument));
}
else if(ops[i].type == 2)
{
int k = Convert.ToInt32(ops[i].argument);
String last = sb.ToString().Substring(sb.Length - k);
sb.Remove(sb.Length - k, k);
operations.Push(new Operation(6, last));
}
else if(ops[i].type == 3)
{
int k = Convert.ToInt32(ops[i].argument);
Console.WriteLine(sb.ToString()[k-1]);
}
else if(ops[i].type == 4)
{
if(operations.Count > 0)
{
Operation operation = operations.Pop();
if(operation.type == 6)
{
sb.Append(operation.argument);
}
else if(operation.type == 5)
{
int k = operation.argument.Length;
sb.Remove(sb.Length - k, k);
}
}
}
}
}
}Problem Solution in Python
class Editor:
def __init__(self):
self.data = []
self.histories = []
def backup(self):
self.histories.append(self.data.copy())
def insert(self, content):
self.backup()
self.data += list(content)
def delete(self, count):
self.backup()
endIdx = len(self.data) - int(count)
self.data = self.data[:endIdx]
def printIdx(self, idx):
print(self.data[int(idx) - 1])
def undo(self, arg = None):
self.data = self.histories.pop()
if not self.data:
self.data = []
def commandToFn(self, command):
return {
"1": "insert",
"2": "delete",
"3": "printIdx",
"4": "undo",
}[command]
def execute(self, instruction):
command = instruction[0]
content = instruction[1] if len(instruction) == 2 else None
return getattr(self, self.commandToFn(command))(content)
if __name__ == '__main__':
editor = Editor()
for op in range(int(input())):
editor.execute(input().split())Problem Solution in JavaScript
function processData(input) {
input = input.split('\n');
let S = "";
let lastStateOfS = [S];
for(let i = 1; i < input.length; i++){
let query = input[i].split(' ');
if(query[0] === '1'){
S += query[1];
lastStateOfS.push(S)
} else if (query[0] === '2'){
let charsToDelete = +query[1];
S = S.substring(0, S.length - charsToDelete);
lastStateOfS.push(S)
} else if (query[0] === '3'){
console.log(S[+query[1] - 1]);
} else{
lastStateOfS.pop();
S = lastStateOfS[lastStateOfS.length - 1];
}
}
return S;
}Solve original Problem on HackerRank here. Checkout more HackerRank Problems