Dada una string S de tamaño N , que consta de dígitos [0 – 9] y el carácter ‘.’ , la tarea es imprimir la string que se puede obtener presionando el teclado móvil en la secuencia dada.
Nota: ‘.’ representa un descanso mientras se escribe.
A continuación se muestra la imagen para representar los caracteres asociados con cada número en el teclado.
Ejemplos:
Entrada: S = «234»
Salida: ADG
Explicación:
Al presionar las teclas 2, 3 y 4 una vez, la string resultante es «ADG».Entrada: S = “22.22”
Salida: BB
Explicación:
Presionar la tecla 2 dos veces da B, y luego presionar nuevamente la tecla dos veces da B. Por lo tanto, la string resultante es “BB”.
Enfoque: el problema dado se puede resolver almacenando las asignaciones de teclados móviles en una array y luego atravesar la string S y convertirla en su string equivalente. Siga los pasos a continuación para resolver el problema:
- Inicialice una string vacía, diga ans para almacenar el resultado requerido.
- Almacena la string asociada a cada tecla del teclado del móvil en un array nums[] de forma que nums[i] representa el conjunto de caracteres al pulsar el dígito i .
- Recorra la string S dada usando la variable i y realice los siguientes pasos:
- Si S[i] es igual a ‘.’ , luego incremente i en 1 y continúe con la siguiente iteración .
- De lo contrario, inicialice una variable cnt como 0 para almacenar el recuento de los mismos caracteres.
- Iterar hasta que S[i] sea igual a S[i + 1] y en cada iteración verificar las siguientes condiciones:
- Si cnt es igual a 2 y S[i] es 2, 3, 4, 5, 6 u 8, salga del ciclo porque las teclas: 2, 3, 4, 5, 6 y 8 contienen el mismo número de caracteres, es decir, 3.
- Si cnt es igual a 3 y S[i] es 7 o 9, salga del bucle porque las teclas: 7 y 9 contienen el mismo número de caracteres, es decir, 4.
- Incremente el valor de cnt e i en 1.
- Si S[i] es 7 o 9, agregue el carácter nums[str[i]][cnt%4] a la string ans .
- De lo contrario, agregue el carácter nums[str[i]][cnt%3] a la string ans .
- Incrementa el valor de i en 1.
- Después de completar los pasos anteriores, imprima la string de valor como resultado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
void printSentence(string str)
{
// Store the mobile keypad mappings
char nums[][5]
= { "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV",
"WXYZ" };
// Traverse the string str
int i = 0;
while (str[i] != '\0') {
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.') {
i++;
continue;
}
// Stores the number of
// continuous clicks
int count = 0;
// Iterate a loop to find the
// count of same characters
while (str[i + 1]
&& str[i] == str[i + 1]) {
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2
&& ((str[i] >= '2'
&& str[i] <= '6')
|| (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3
&& (str[i] == '7'
|| str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (str[i] == '\0')
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9') {
cout << nums[str[i] - 48][count % 4];
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else {
cout << nums[str[i] - 48][count % 3];
}
i++;
}
}
// Driver Code
int main()
{
string str = "234";
printSentence(str);
return 0;
}
Java
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
static void printSentence(String S)
{
// Store the mobile keypad mappings
String nums[]
= { "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV", "WXYZ" };
char str[] = S.toCharArray();
// Traverse the string str
int i = 0;
while (i < str.length) {
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.') {
i++;
continue;
}
// Stores the number of
// continuous clicks
int count = 0;
// Iterate a loop to find the
// count of same characters
while (i + 1 < str.length
&& str[i] == str[i + 1]) {
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2
&& ((str[i] >= '2' && str[i] <= '6')
|| (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3
&& (str[i] == '7'
|| str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (i == str.length)
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9') {
System.out.print(
nums[str[i] - 48].charAt(count % 4));
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else {
System.out.print(
nums[str[i] - 48].charAt(count % 3));
}
i++;
}
}
// Driver Code
public static void main(String[] args)
{
String str = "234";
printSentence(str);
}
}
// This code is contributed by Kingash.
Python3
# Python3 program for the above approach # Function to convert mobile numeric # keypad sequence into its equivalent # string def printSentence(str1): # Store the mobile keypad mappings nums = [ "", "", "ABC", "DEF", "GHI", "JKL", "MNO", "PQRS", "TUV", "WXYZ" ] # Traverse the string str1 i = 0 while (i < len(str1)): # If the current character is # '.', then continue to the # next iteration if (str1[i] == '.'): i += 1 continue # Stores the number of # continuous clicks count = 0 # Iterate a loop to find the # count of same characters while (i + 1 < len(str1) and str1[i + 1] and str1[i] == str1[i + 1]): # 2, 3, 4, 5, 6 and 8 keys will # have maximum of 3 letters if (count == 2 and ((str1[i] >= '2' and str1[i] <= '6') or (str1[i] == '8'))): break # 7 and 9 keys will have # maximum of 4 keys elif (count == 3 and (str1[i] == '7' or str1[i] == '9')): break count += 1 i += 1 # Handle the end condition if (i < len(str)): break # Check if the current pressed # key is 7 or 9 if (str1[i] == '7' or str1[i] == '9'): print(nums[ord(str1[i]) - 48][count % 4], end = "") # Else, the key pressed is # either 2, 3, 4, 5, 6 or 8 else: print(nums[ord(str1[i]) - 48][count % 3], end = "") i += 1 # Driver Code if __name__ == '__main__': str1 = "234" printSentence(str1) # This code is contributed by bgangwar59
C#
// C# program for the above approach
using System;
public class GFG
{
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
static void printSentence(string S)
{
// Store the mobile keypad mappings
string[] nums
= { "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV", "WXYZ" };
char[] str = S.ToCharArray();
// Traverse the string str
int i = 0;
while (i < str.Length) {
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.') {
i++;
continue;
}
// Stores the number of
// continuous clicks
int count = 0;
// Iterate a loop to find the
// count of same characters
while (i + 1 < str.Length
&& str[i] == str[i + 1]) {
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2
&& ((str[i] >= '2' && str[i] <= '6')
|| (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3
&& (str[i] == '7'
|| str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (i == str.Length)
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9') {
Console.Write(nums[str[i] - 48][count % 4]);
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else {
Console.Write(nums[str[i] - 48][count % 3]);
}
i++;
}
}
// Driver Code
public static void Main(string[] args)
{
string str = "234";
printSentence(str);
}
}
// This code is contributed by ukasp.
Javascript
<script>
// Javascript program for the above approach
// Function to convert mobile numeric
// keypad sequence into its equivalent
// string
function printSentence(S)
{
// Store the mobile keypad mappings
let nums = [ "", "", "ABC", "DEF", "GHI",
"JKL", "MNO", "PQRS", "TUV", "WXYZ"];
let str = S.split("");
// Traverse the string str
let i = 0;
while (i < str.length)
{
// If the current character is
// '.', then continue to the
// next iteration
if (str[i] == '.')
{
i++;
continue;
}
// Stores the number of
// continuous clicks
let count = 0;
// Iterate a loop to find the
// count of same characters
while (i + 1 < str.length &&
str[i] == str[i + 1])
{
// 2, 3, 4, 5, 6 and 8 keys will
// have maximum of 3 letters
if (count == 2 && ((str[i] >= '2' &&
str[i] <= '6') || (str[i] == '8')))
break;
// 7 and 9 keys will have
// maximum of 4 keys
else if (count == 3 && (str[i] == '7' ||
str[i] == '9'))
break;
count++;
i++;
// Handle the end condition
if (i == str.length)
break;
}
// Check if the current pressed
// key is 7 or 9
if (str[i] == '7' || str[i] == '9')
{
document.write(
nums[str[i].charCodeAt(0) - 48][count % 4]);
}
// Else, the key pressed is
// either 2, 3, 4, 5, 6 or 8
else
{
document.write(
nums[str[i].charCodeAt(0) - 48][count % 3]);
}
i++;
}
}
// Driver Code
let str = "234";
printSentence(str);
// This code is contributed by _saurabh_jaiswal.
</script>
Producción:
ADG
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por avllikhita y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA