Program to print the given digit in words

Given a number N, the task is to convert every digit of the number into words.

Examples:

Input: N = 1234 
Output: One Two Three Four 
Explanation:
Every digit of the given number has been converted into its corresponding word.

Input: N = 567 
Output: Five Six Seven 

Approach: The idea is to traverse through every digit of the number and use switch-case. Since there are only ten possible values for digits, ten cases can be defined inside a switch block. For each digit, its corresponding case block will be executed and that digit will get printed in words.

Below is the implementation of the above approach: 

CPP

// C++ implementation of the above approach

#include "bits/stdc++.h"
using namespace std;

// Function to return the word
// of the corresponding digit
void printValue(char digit)
{

    // Switch block to check for each digit c
    switch (digit) {

    // For digit 0
    case '0':
        cout << "Zero ";
        break;

    // For digit 1
    case '1':
        cout << "One ";
        break;

    // For digit 2
    case '2':
        cout << "Two ";
        break;

    // For digit 3
    case '3':
        cout << "Three ";
        break;

    // For digit 4
    case '4':
        cout << "Four ";
        break;

    // For digit 5
    case '5':
        cout << "Five ";
        break;

    // For digit 6
    case '6':
        cout << "Six ";
        break;

    // For digit 7
    case '7':
        cout << "Seven ";
        break;

    // For digit 8
    case '8':
        cout << "Eight ";
        break;

    // For digit 9
    case '9':
        cout << "Nine ";
        break;
    }
}

// Function to iterate through every
// digit in the given number
void printWord(string N)
{
    int i, length = N.length();

    // Finding each digit of the number
    for (i = 0; i < length; i++) {

        // Print the digit in words
        printValue(N[i]);
    }
}

// Driver code
int main()
{
    string N = "123";
    printWord(N);
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

// Function to return the word
// of the corresponding digit
static void printValue(char digit)
{

    // Switch block to check for each digit c
    switch (digit)
    {

    // For digit 0
    case '0':
        System.out.print("Zero ");
        break;

    // For digit 1
    case '1':
        System.out.print("One ");
        break;

    // For digit 2
    case '2':
        System.out.print("Two ");
        break;

    // For digit 3
    case '3':
        System.out.print("Three ");
        break;

    // For digit 4
    case '4':
        System.out.print("Four ");
        break;

    // For digit 5
    case '5':
        System.out.print("Five ");
        break;

    // For digit 6
    case '6':
        System.out.print("Six ");
        break;

    // For digit 7
    case '7':
        System.out.print("Seven ");
        break;

    // For digit 8
    case '8':
        System.out.print("Eight ");
        break;

    // For digit 9
    case '9':
        System.out.print("Nine ");
        break;
    }
}

// Function to iterate through every
// digit in the given number
static void printWord(String N)
{
    int i, length = N.length();

    // Finding each digit of the number
    for (i = 0; i < length; i++)
    {

        // Print the digit in words
        printValue(N.charAt(i));
    }
}

// Driver code
public static void main(String[] args)
{
    String N = "123";
    printWord(N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach

# Function to return the word
# of the corresponding digit
def printValue(digit):

    # Switch block to check for each digit c

    # For digit 0
    if digit == '0':
        print("Zero ", end = " ")

    # For digit 1
    elif digit == '1':
        print("One ", end = " ")

    # For digit 2
    elif digit == '2':
        print("Two ", end = " ")

    #For digit 3
    elif digit=='3':
        print("Three",end=" ")

    # For digit 4
    elif digit == '4':
        print("Four ", end = " ")

    # For digit 5
    elif digit == '5':
        print("Five ", end = " ")

    # For digit 6
    elif digit == '6':
        print("Six ", end = " ")

    # For digit 7
    elif digit == '7':
        print("Seven", end = " ")

    # For digit 8
    elif digit == '8':
        print("Eight", end = " ")

    # For digit 9
    elif digit == '9':
        print("Nine ", end = " ")

# Function to iterate through every
# digit in the given number
def printWord(N):
    i = 0
    length = len(N)

    # Finding each digit of the number
    while i < length:

        # Print the digit in words
        printValue(N[i])
        i += 1

# Driver code
N = "123"
printWord(N)

# This code is contributed by mohit kumar 29

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the word
    // of the corresponding digit
    static void printValue(char digit)
    {

        // Switch block to check for each digit c
        switch (digit)
        {

        // For digit 0
        case '0':
            Console.Write("Zero ");
            break;

        // For digit 1
        case '1':
            Console.Write("One ");
            break;

        // For digit 2
        case '2':
            Console.Write("Two ");
            break;

        // For digit 3
        case '3':
            Console.Write("Three ");
            break;

        // For digit 4
        case '4':
            Console.Write("Four ");
            break;

        // For digit 5
        case '5':
            Console.Write("Five ");
            break;

        // For digit 6
        case '6':
            Console.Write("Six ");
            break;

        // For digit 7
        case '7':
            Console.Write("Seven ");
            break;

        // For digit 8
        case '8':
            Console.Write("Eight ");
            break;

        // For digit 9
        case '9':
            Console.Write("Nine ");
            break;
        }
    }

    // Function to iterate through every
    // digit in the given number
    static void printWord(string N)
    {
        int i, length = N.Length;

        // Finding each digit of the number
        for (i = 0; i < length; i++)
        {

            // Print the digit in words
            printValue(N[i]);
        }
    }

    // Driver code
    public static void Main()
    {
        string N = "123";
        printWord(N);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

      // JavaScript implementation of
      // the above approach

      // Function to return the word
      // of the corresponding digit
      function printValue(digit) {
        // Switch block to check for
        // each digit c
        switch (digit) {
          // For digit 0
          case "0":
            document.write("Zero ");
            break;

          // For digit 1
          case "1":
            document.write("One ");
            break;

          // For digit 2
          case "2":
            document.write("Two ");
            break;

          // For digit 3
          case "3":
            document.write("Three ");
            break;

          // For digit 4
          case "4":
            document.write("Four ");
            break;

          // For digit 5
          case "5":
            document.write("Five ");
            break;

          // For digit 6
          case "6":
            document.write("Six ");
            break;

          // For digit 7
          case "7":
            document.write("Seven ");
            break;

          // For digit 8
          case "8":
            document.write("Eight ");
            break;

          // For digit 9
          case "9":
            document.write("Nine ");
            break;
        }
      }

      // Function to iterate through every
      // digit in the given number
      function printWord(N) {
        var i,
          length = N.length;

        // Finding each digit of the number
        for (i = 0; i < length; i++) {
          // Print the digit in words
          printValue(N[i]);
        }
      }

      // Driver code
      var N = "123";
      printWord(N);

</script>

Output

One Two Three 

Time Complexity: O(L), Here L is the length of the string
Auxiliary Space: O(1), As constant extra space is used.

Recursive Approach

The idea is to recursively call the function till the number becomes zero by dividing it by 10 and storing the remainder in a variable. Then we print the digit from the string array as the digit will store the index of the number to be printed from the array.  

we print the number after the recursive call to maintain the order of digits in the input number, if we print before the recursive function call the digit name will be printed in reversed order. Since we are dividing n by 10 in each recursive call the recurrence relation will be  T(n) = T(n/10) + 1

Below is the implementation of the above approach: 

C++

//C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

void ToDigits(int n, string arr[])
{

    // base case
    if (n == 0) {
        return;
    }

    // storing the last digit of the number and updating
    // number
    int digit = n % 10;
    n = n / 10;

    // recursive call
    ToDigits(n, arr);

    // printing the digits form the string array storing name
    // of the given index
    cout << arr[digit] << " ";
}

int main()
{

    string arr[10]
        = { "zero", "one", "two",   "three", "four",
            "five", "six", "seven", "eight", "nine" };
    int n;
    n = 123;  //it can be changed to take user input

    ToDigits(n, arr);

    return 0;
}
//This code is contributed by rahulpatel43433

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
  static void ToDigits(int n, String[] arr)
  {

    // base case
    if (n == 0) {
      return;
    }

    // storing the last digit of the number and updating
    // number
    int digit = n % 10;
    n = n / 10;

    // recursive call
    ToDigits(n, arr);

    // printing the digits form the string array storing name
    // of the given index
    System.out.print(arr[digit]);
    System.out.print(" ");
  }

  // Driver Code
  public static void main(String args[])
  {
    String[] arr = { "zero", "one", "two",   "three", "four",    "five", "six", "seven", "eight", "nine" };
    int n = 123;  //it can be changed to take user input

    ToDigits(n, arr);
  }
}

// This code is contributed by shinjanpatra

Python3

# Python implementation of above approach
def ToDigits(n, arr):

    # base case
    if (n == 0):
        return

    # storing the last digit of the number and updating
    # number
    digit = n % 10
    n = n // 10

    # recursive call
    ToDigits(n, arr)

    # printing the digits form the string array storing name
    # of the given index
    print(arr[digit] , end = " ")


# driver code
arr = [ "zero", "one", "two", "three", "four",
            "five", "six", "seven", "eight", "nine" ]
n = 123 #it can be changed to take user input

ToDigits(n, arr)

# This code is contributed by shinjanpatra

C#

//c# implementation of above approach
using System;

public class GFG {

    static void ToDigits(int n, String[] arr)
    {
        // base case
        if (n == 0) {
            return;
        }

        // storing the last digit of the number and updating
        // number
        int digit = n % 10;
        n = n / 10;

        // recursive call
        ToDigits(n, arr);

        // printing the digits form the string array storing name
        // of the given index
        Console.Write(arr[digit]+" ");
    }

    // Driver program to test above
    public static void Main()
    {
        String[] arr = new string[10]{ "zero", "one", "two",   "three", "four", "five",
        "six", "seven", "eight", "nine" };

        int n;
        n = 123;  //it can be changed to take user input

        ToDigits(n, arr);
    }
}
// This code is contributed by aditya942003patil

Javascript

<script>

// JavaScript implementation of above approach
function ToDigits(n, arr)
{

    // base case
    if (n == 0) {
        return;
    }

    // storing the last digit of the number and updating
    // number
    let digit = n % 10;
    n = Math.floor(n / 10);

    // recursive call
    ToDigits(n, arr);

    // printing the digits form the string array storing name
    // of the given index
    document.write(arr[digit] , " ");
}

// driver code
let arr = [ "zero", "one", "two", "three", "four",
            "five", "six", "seven", "eight", "nine" ]
let n = 123; //it can be changed to take user input

ToDigits(n, arr);

// This code is contributed by shinjanpatra

</script>

Output

one two three 

Time Complexity: O(log₁₀ n)
Auxiliary Space: O(log₁₀ n) for recursive call stack

Approach 3 : By using Python Dictionary. 

Algorithm –

1. Create a dictionary named digits, assign key-value pairs of numbers to words respectively, as shown in code below.
2. Iterate the string character one by one and print the value of the dictionary, key associated with it.

C++

#include <iostream>
#include <string>
#include <map>

using namespace std;

// Function to iterate through every digit in the given number
void printWord(string N) {
    map<char, string> digits {
        {'1', "One"}, {'2', "Two"}, {'3', "Three"}, {'4', "Four"},
        {'5', "Five"}, {'6', "Six"}, {'7', "Seven"}, {'8', "Eight"},
        {'9', "Nine"}, {'0', "Zero"}
    };

    for (char number : N) {
        cout << digits[number] << " ";
    }
}

// Driver code
int main() {
    string N = "123";
    printWord(N);
    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;

public class Main {
    // Function to iterate through every digit in the given number
    public static void printWord(String N) {
        Map<Character, String> digits = new HashMap<Character, String>() {{
            put('1', "One");
            put('2', "Two");
            put('3', "Three");
            put('4', "Four");
            put('5', "Five");
            put('6', "Six");
            put('7', "Seven");
            put('8', "Eight");
            put('9', "Nine");
            put('0', "Zero");
        }};

        for (char number : N.toCharArray()) {
            System.out.print(digits.get(number) + " ");
        }
    }

    // Driver code
    public static void main(String[] args) {
        String N = "123";
        printWord(N);
    }
}

Python3

# Python3 Code implementation using dictionary

# Function to iterate through every
# digit in the given number
def printWord(N):
    digits = {'1':'One','2':'Two','3':'Three','4':'Four','5':'Five','6':'Six','7':'Seven','8':'Eight','9':'Nine','0':'Zero'}
    for number in N:
        print(digits[number] , end = ' ')

# Driver code
N = "123"
printWord(N)

# This code is contributed by Pratik Gupta (guptapratik)

C#

using System;
using System.Collections.Generic;

class Program {
    static void Main(string[] args)
    {
        string N = "123";
        PrintWord(N);
    }

    // Function to iterate through every digit in the given
    // number
    static void PrintWord(string N)
    {
        // Dictionary to store the word representation of
        // each digit
        Dictionary<char, string> digits
            = new Dictionary<char, string>{
                  { '1', "One" },   { '2', "Two" },
                  { '3', "Three" }, { '4', "Four" },
                  { '5', "Five" },  { '6', "Six" },
                  { '7', "Seven" }, { '8', "Eight" },
                  { '9', "Nine" },  { '0', "Zero" }
              };

        // Iterate through each digit in the number and
        // print its word representation
        foreach(char number in N)
        {
            Console.Write(digits[number] + " ");
        }
    }
}

Javascript

// Function to iterate through every
// digit in the given number
function printWord(N) {
    const digits = {'1':'One','2':'Two','3':'Three','4':'Four','5':'Five','6':'Six','7':'Seven','8':'Eight','9':'Nine','0':'Zero'};
    for (let number of N) {
        console.log(digits[number] + ' ');
    }
}

// Driver code
const N = "123";
printWord(N);

Output

One Two Three 

Time Complexity: O(n), where n is the length of string
Auxiliary Space: O(1)