Dada la string str del tipo “3(ab)4(cd)” , la tarea es expandirla a “abababcdcdcdcd”, donde los números enteros son del rango [1, 9] .
Este problema se planteó en una entrevista de ThoughtWorks realizada en octubre de 2018.
Ejemplos:
Entrada: str = “3(ab)4(cd)”
Salida: abababcdcdcdcd
Entrada: str = “2(kl)3(ap)”
Salida: klklapapap
Enfoque: Atravesamos la string y esperamos un valor numérico, num , para aparecer en la posición i . Tan pronto como llega, buscamos en i + 1 un ‘(‘ . Si está presente, el programa entra en un ciclo para extraer lo que esté dentro de ‘(‘ y ‘)’ y concatenarlo en una string vacía, temp . Más tarde, otro ciclo imprime la string generada num número de veces.Repita estos pasos hasta que la string termine.
A continuación se muestra la implementación del enfoque:
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to expand and print the given string
void expandString(string strin)
{
string temp = "";
int j;
for (int i = 0; i < strin.length(); i++)
{
if (strin[i] >= 0)
{
// Subtract '0' to convert char to int
int num = strin[i] - '0';
if (strin[i + 1] == '(')
{
// Characters within brackets
for (j = i + 1; strin[j] != ')'; j++)
{
if ((strin[j] >= 'a' && strin[j] <= 'z') ||
(strin[j] >= 'A' && strin[j] <= 'Z'))
{
temp += strin[j];
}
}
// Expanding
for (int k = 1; k <= num; k++)
{
cout << (temp);
}
// Reset the variables
num = 0;
temp = "";
if (j < strin.length())
{
i = j;
}
}
}
}
}
// Driver code
int main()
{
string strin = "3(ab)4(cd)";
expandString(strin);
}
// This code is contributed by Surendra_Gangwar
Java
// Java implementation of the approach
public class GFG {
// Function to expand and print the given string
static void expandString(String strin)
{
String temp = "";
int j;
for (int i = 0; i < strin.length(); i++) {
if (strin.charAt(i) >= 0) {
// Subtract '0' to convert char to int
int num = strin.charAt(i) - '0';
if (strin.charAt(i + 1) == '(') {
// Characters within brackets
for (j = i + 1; strin.charAt(j) != ')'; j++) {
if ((strin.charAt(j) >= 'a'
&& strin.charAt(j) <= 'z')
|| (strin.charAt(j) >= 'A'
&& strin.charAt(j) <= 'Z')) {
temp += strin.charAt(j);
}
}
// Expanding
for (int k = 1; k <= num; k++) {
System.out.print(temp);
}
// Reset the variables
num = 0;
temp = "";
if (j < strin.length()) {
i = j;
}
}
}
}
}
// Driver code
public static void main(String args[])
{
String strin = "3(ab)4(cd)";
expandString(strin);
}
}
Python3
# Python3 implementation of the approach
# Function to expand and print the given string
def expandString(strin):
temp = ""
j = 0
i = 0
while(i < len(strin)):
if (strin[i] >= "0"):
# Subtract '0' to convert char to int
num = ord(strin[i])-ord("0")
if (strin[i + 1] == '('):
# Characters within brackets
j = i + 1
while(strin[j] != ')'):
if ((strin[j] >= 'a' and strin[j] <= 'z') or \
(strin[j] >= 'A' and strin[j] <= 'Z')):
temp += strin[j]
j += 1
# Expanding
for k in range(1, num + 1):
print(temp,end="")
# Reset the variables
num = 0
temp = ""
if (j < len(strin)):
i = j
i += 1
# Driver code
strin = "3(ab)4(cd)"
expandString(strin)
# This code is contributed by shubhamsingh10
C#
// C# implementation of
// the above approach
using System;
class GFG{
// Function to expand and
// print the given string
static void expandString(string strin)
{
string temp = "";
int j;
for (int i = 0;
i < strin.Length; i++)
{
if (strin[i] >= 0)
{
// Subtract '0' to
// convert char to int
int num = strin[i] - '0';
if (strin[i + 1] == '(')
{
// Characters within brackets
for (j = i + 1;
strin[j] != ')'; j++)
{
if ((strin[j] >= 'a' &&
strin[j] <= 'z') ||
(strin[j] >= 'A' &&
strin[j] <= 'Z'))
{
temp += strin[j];
}
}
// Expanding
for (int k = 1; k <= num; k++)
{
Console.Write(temp);
}
// Reset the variables
num = 0;
temp = "";
if (j < strin.Length)
{
i = j;
}
}
}
}
}
// Driver code
public static void Main(String [] args)
{
string strin = "3(ab)4(cd)";
expandString(strin);
}
}
// This code is contributed by Chitranayal
Javascript
<script>
// Javascript implementation of the approach
// Function to expand and print the given string
function expandString(strin)
{
let temp = "";
let j;
for (let i = 0; i < strin.length; i++) {
if (strin[i].charCodeAt(0) >= 0) {
// Subtract '0' to convert char to int
let num = strin[i].charCodeAt(0) - '0'.charCodeAt(0);
if (strin[i+1] == '(') {
// Characters within brackets
for (j = i + 1; strin[j] != ')'; j++) {
if ((strin[j] >= 'a'
&& strin[j] <= 'z')
|| (strin[j] >= 'A'
&& strin[j] <= 'Z')) {
temp += strin[j];
}
}
// Expanding
for (let k = 1; k <= num; k++) {
document.write(temp);
}
// Reset the variables
num = 0;
temp = "";
if (j < strin.length) {
i = j;
}
}
}
}
}
// Driver code
let strin = "3(ab)4(cd)";
expandString(strin);
// This code is contributed by rag2127
</script>
Producción:
abababcdcdcdcd
Complejidad temporal: O(N*N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por thecoducer y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA