Imprime dígitos individuales como palabras sin usar if o switch

Dado un número, imprime palabras para dígitos individuales. No está permitido usar if o switch .
Ejemplos: 

Input:  n = 123
Output: One Two Three

Input:  n = 350
Output: Three Five Zero

Le recomendamos encarecidamente que minimice su navegador y que pruebe esto usted mismo primero.  
La idea es utilizar una array de strings para almacenar asignaciones de dígitos a palabras. A continuación se muestran los pasos.
Deje que el número de entrada sea n. 

1. Cree una array de strings para almacenar la asignación de dígitos a palabras.
2. Cree otra array digits[] para almacenar dígitos individuales de n.
3. Recorra los dígitos de n y guárdelos en digits[]. Tenga en cuenta que la forma estándar de recorrido mediante el almacenamiento repetido de n%10 y haciendo n = n/10, atraviesa los dígitos en orden inverso.
4. Recorra la array de dígitos desde el final hasta el principio e imprima palabras usando el mapeo creado en el paso 1.

A continuación se muestra la implementación de la idea anterior. 

// C++ program to print individual words without if and
// without switch
#include <bits/stdc++.h>
using namespace std;
 
// To store digit to word mapping
char word[][10] = {"zero", "one", "two", "three","four",
                   "five", "six", "seven", "eight", "nine"};
 
void printWordsWithoutIfSwitch(int n)
{
    // Store individual digits
    int digits[10]; // a 32 bit int has at-most 10 digits
 
    int dc = 0; // Initialize digit count for given number 'n'
 
    // The below loop stores individual digits of n in
    // reverse order. do-while is used to handle "0" input
    do
    {
        digits[dc] = n%10;
        n = n/10;
        dc++;
    } while (n != 0);
 
    // Traverse individual digits and print words using
    // word[][]
    for (int i=dc-1; i>=0; i--)
       cout << word[digits[i]] << " ";
}
 
// Driver program
int main()
{
    int n = 350;
    printWordsWithoutIfSwitch(n);
    return 0;
}

// Java program to print individual words without
//  if and without switch
class GFG
{
 
// To store digit to word mapping
static String word[] = {"zero", "one", "two", "three","four",
                "five", "six", "seven", "eight", "nine"};
 
static void printWordsWithoutIfSwitch(int n)
{
    // Store individual digits
    int digits[] = new int[10]; // a 32 bit int has at-most 10 digits
 
    int dc = 0; // Initialize digit count for given number 'n'
 
    // The below loop stores individual digits of n in
    // reverse order. do-while is used to handle "0" input
    do
    {
        digits[dc] = n % 10;
        n = n/10;
        dc++;
    } while (n != 0);
 
    // Traverse individual digits and print words using
    // word[][]
    for (int i = dc - 1; i >= 0; i--)
        System.out.print(word[digits[i]] + " ");
}
 
// Driver program
public static void main(String[] args)
{
    int n = 350;
    printWordsWithoutIfSwitch(n);
}
}
 
// This code has been contributed by 29AjayKumar

// C# program to print individual words without
// if and without switch
using System;
 
class GFG
{
 
// To store digit to word mapping
static String []word = {"zero", "one", "two", "three","four",
                "five", "six", "seven", "eight", "nine"};
 
static void printWordsWithoutIfSwitch(int n)
{
    // Store individual digits
    int []digits = new int[10]; // a 32 bit int has at-most 10 digits
 
    int dc = 0; // Initialize digit count for given number 'n'
 
    // The below loop stores individual digits of n in
    // reverse order. do-while is used to handle "0" input
    do
    {
        digits[dc] = n % 10;
        n = n/10;
        dc++;
    } while (n != 0);
 
    // Traverse individual digits and print words using
    // word[][]
    for (int i = dc - 1; i >= 0; i--)
        Console.Write(word[digits[i]] + " ");
}
 
// Driver program
public static void Main(String[] args)
{
    int n = 350;
    printWordsWithoutIfSwitch(n);
}
}
 
// This code contributed by Rajput-Ji

# Python program to print individual words without if and
# without switch
 
# To store digit to word mapping
word= ["zero", "one", "two", "three","four","five",
                "six", "seven", "eight", "nine"]
 
def printWordsWithoutIfSwitch(n):
 
    # Store individual digits
    digits = [0 for i in range(10)] # a 32 bit has at-most 10 digits
 
    dc = 0 # Initialize digit count for given number 'n'
 
    # The below loop stores individual digits of n in
    # reverse order. do-while is used to handle "0" input
    while True:
        digits[dc] = n%10
        n = n//10
        dc += 1
        if(n==0):
            break
 
    # Traverse individual digits and print words using
    # word[][]
    for i in range(dc-1,-1,-1):
        print(word[digits[i]],end=" ")
 
# Driver program
n = 350
printWordsWithoutIfSwitch(n)
 
# This code is contributed by mohit kumar 29

<script>
// Javascript program to print individual words without
//  if and without switch
     
    // To store digit to word mapping
    let word=["zero", "one", "two", "three","four",
                "five", "six", "seven", "eight", "nine"];
     
    function printWordsWithoutIfSwitch(n)
    {
     
        // Store individual digits
    let digits = new Array(10); // a 32 bit int has at-most 10 digits
   
    let dc = 0; // Initialize digit count for given number 'n'
   
    // The below loop stores individual digits of n in
    // reverse order. do-while is used to handle "0" input
    do
    {
        digits[dc] = n % 10;
        n = Math.floor(n/10);
        dc++;
    } while (n != 0);
   
    // Traverse individual digits and print words using
    // word[][]
    for (let i = dc - 1; i >= 0; i--)
        document.write(word[digits[i]] + " ");
    }
     
    // Driver program  
    let n = 350;
    printWordsWithoutIfSwitch(n);  
     
    // This code is contributed by unknown2108
</script>

three five zero 

Complejidad de tiempo: O(|n|), donde |n| es el número de dígitos en el número dado n.
Espacio auxiliar: O (1), para almacenar el dígito a la asignación de palabras

Gracias a Utkarsh Trivedi por sugerir la solución anterior.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.