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 |
One Two Three
Related Article: Print individual digits as words without using if or switch
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