Given an integer N, the task is to convert the given number into words.
Examples :
Input: N = 438237764
Output: Four Hundred Thirty Eight Million Two Hundred Thirty Seven Thousand Seven Hundred Sixty FourInput: N = 1000
Output: One Thousand
Approach: This approach is valid for converting number into words for up to 20 digits, and is basically based on Brute Force Logic.
The idea is to break down the number into International Number System, i.e., smaller groups of three digits (hundreds, tens, and ones), and convert each group into words.
Consider the below illustration:
Suppose N = 1234567
Now, process this number N in groups of three digits from right to left.
Iteration 1:
- Current Group: 567
- Word Representation: Five Hundred Sixty-Seven
- Multiplier: “”
At this point, the partial word representation is “Five Hundred Sixty-Seven”.
Iteration 2:
- Current Group: 234
- Word Representation: Two Hundred Thirty-Four
- Multiplier: Thousand
The updated word representation becomes “Two Hundred Thirty-Four Thousand Five Hundred Sixty-Seven”.
Iteration 3:
- Current Group: 1
- English Representation: One
- Multiplier: Million
Hence, finally the given number into words becomes “One Million Two Hundred Thirty-Four Thousand Five Hundred Sixty-Seven“.
NOTE: The following code supports numbers up to 15 digits, i.e., numbers from 0 to trillions. The code prints according to the western number system.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;string numberToWords(long long int n){ long long int limit = 1000000000000, curr_hun, t = 0; // If zero return zero if (n == 0) return ("Zero"); // Array to store the powers of 10 string multiplier[] = { "", "Trillion", "Billion", "Million", "Thousand" }; // Array to store numbers till 20 string first_twenty[] = { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }; // Array to store multiples of ten string tens[] = { "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" }; // If number is less than 20, return without any further // computation if (n < 20) return (first_twenty[n]); // Answer variable to store the conversion string answer = ""; for (long long int i = n; i > 0; i %= limit, limit /= 1000) { // Store the value in multiplier[t], i.e n = // 1000000, then r = 1, for multiplier(million), 0 // for multipliers(trillion and billion) curr_hun = i / limit; // It might be possible that the current multiplier // is bigger than your number while (curr_hun == 0) { // Set i as the remainder obtained when n was // divided by the limit i %= limit; // Divide the limit by 1000, shifts the // multiplier limit /= 1000; // Get the current value in hundreds, as // English system works in hundreds curr_hun = i / limit; // Shift the multiplier ++t; } // If current hundred is greater than 99, Add the // hundreds' place if (curr_hun > 99) answer += (first_twenty[curr_hun / 100] + " Hundred "); // Bring the current hundred to tens curr_hun = curr_hun % 100; // If the value in tens belongs to [1,19], add using // the first_twenty if (curr_hun > 0 && curr_hun < 20) answer += (first_twenty[curr_hun] + " "); // If curr_hun is now a multiple of 10, but not 0 // Add the tens' value using the tens array else if (curr_hun % 10 == 0 && curr_hun != 0) answer += (tens[curr_hun / 10 - 1] + " "); // If the value belongs to [21,99], excluding the // multiples of 10 Get the ten's place and one's // place, and print using the first_twenty array else if (curr_hun > 20 && curr_hun < 100) answer += (tens[curr_hun / 10 - 1] + " " + first_twenty[curr_hun % 10] + " "); // If Multiplier has not become less than 1000, // shift it if (t < 4) answer += (multiplier[++t] + " "); } return (answer);}// Driver code.int main(){ // Input 1 long long int n = 36; cout << numberToWords(n) << '\n'; // Input 2 n = 123456789; cout << numberToWords(n) << '\n'; // Input 3 n = 10101010110001; cout << numberToWords(n) << '\n'; // Input 4 n = 999999999; cout << numberToWords(n) << '\n'; return 0;}// This Code is contributed by Alok Khansali |
Java
public class GFG { static String numberToWords(long n) { long limit = 1000000000000L, curr_hun, t = 0; // If zero return zero if (n == 0) return ("Zero"); // Array to store the powers of 10 String multiplier[] = { "", "Trillion", "Billion", "Million", "Thousand" }; // Array to store numbers till 20 String first_twenty[] = { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }; // Array to store multiples of ten String tens[] = { "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" }; // If number is less than 20, return without any // further computation if (n < 20L) return (first_twenty[(int)n]); String answer = ""; for (long i = n; i > 0; i %= limit, limit /= 1000) { // Store the value in multiplier[t], i.e n = // 1000000, then r = 1, for multiplier(million), // 0 for multipliers(trillion and billion) // multiplier here refers to the current // accessible limit curr_hun = i / limit; // It might be possible that the current // multiplier is bigger than your number while (curr_hun == 0) { // Set i as the remainder obtained when n // was divided by the limit i %= limit; // Divide the limit by 1000, shifts the // multiplier limit /= 1000; // Get the current value in hundreds, as // English system works in hundreds curr_hun = i / limit; // Shift the multiplier ++t; } // If current hundred is greater than 99, Add // the hundreds' place if (curr_hun > 99) answer += (first_twenty[(int)curr_hun / 100] + " Hundred "); // Bring the current hundred to tens curr_hun = curr_hun % 100; // If the value in tens belongs to [1,19], add // using the first_twenty if (curr_hun > 0 && curr_hun < 20) answer += (first_twenty[(int)curr_hun] + " "); // If curr_hun is now a multiple of 10, but not // 0 Add the tens' value using the tens array else if (curr_hun % 10 == 0 && curr_hun != 0) answer += (tens[(int)curr_hun / 10 - 1] + " "); // If the value belongs to [21,99], excluding // the multiples of 10 Get the ten's place and // one's place, and print using the first_twenty // array else if (curr_hun > 20 && curr_hun < 100) answer += (tens[(int)curr_hun / 10 - 1] + " " + first_twenty[(int)curr_hun % 10] + " "); // If Multiplier has not become less than 1000, // shift it if (t < 4) answer += (multiplier[(int)++t] + " "); } return (answer); } // Driver code public static void main(String args[]) { // Input 1 long n = 36L; System.out.println(numberToWords(n)); // Input 2 n = 123456789; System.out.println(numberToWords(n)); // Input 3 n = 10101010110001L; System.out.println(numberToWords(n)); // Input 4 n = 999999999; System.out.println(numberToWords(n)); }}// This Code is contributed by Alok Khansali |
Python
def numberToWords(n): limit, t = 1000000000000, 0 # If zero print zero if (n == 0): print("zero") return # Array to store the powers of 10 multiplier = ["", "Trillion", "Billion", "Million", "Thousand"] # Array to store numbers till 20 first_twenty = ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] # Array to store multiples of ten tens = ["", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] # If number is less than 20, print without any if (n < 20): print(first_twenty[n]) return answer = "" i = n while(i > 0): ''' Store the value in multiplier[t], i.e n = 1000000, then r = 1, for multiplier(million), 0 for multipliers(trillion and billion) multiplier here refers to the current accessible limit ''' curr_hun = i // limit # It might be possible that the current multiplier is bigger than your number while (curr_hun == 0): # Set i as the remainder obtained when n was divided by the limit i %= limit # Divide the limit by 1000, shifts the multiplier limit /= 1000 # Get the current value in hundreds, as English system works in hundreds curr_hun = i // limit # Shift the multiplier t += 1 # If current hundred is greater than 99, Add the hundreds' place if (curr_hun > 99): answer += (first_twenty[curr_hun // 100] + " tensundred ") # Bring the current hundred to tens curr_hun = curr_hun % 100 # If the value in tens belongs to [1,19], add using the first_twenty if (curr_hun > 0 and curr_hun < 20): answer += (first_twenty[curr_hun] + " ") # If curr_hun is now a multiple of 10, but not 0 # Add the tens' value using the tens array elif (curr_hun % 10 == 0 and curr_hun != 0): answer += (tens[(curr_hun//10) - 1] + " ") # If the value belongs to [21,99], excluding the multiples of 10 # Get the ten's place and one's place, and print using the first_twenty array elif (curr_hun > 19 and curr_hun < 100): answer += (tens[(curr_hun//10) - 1] + " " + first_twenty[curr_hun % 10] + " ") # If Multiplier has not become less than 1000, shift it if (t < 4): answer += (multiplier[t] + " ") i = i % limit limit = limit // 1000 print(answer)# Input 1n = 36numberToWords(n)# Input 2n = 123456789numberToWords(n)# Input 3n = 10101010110001numberToWords(n)# Input 4n = 999999999numberToWords(n)# This code is contributed by Alok Khansali |
C#
// Include namespace systemusing System;public class numtowords { public static String numberToWords(long n) { var limit = 1000000000000L; long curr_hun; var t = 0; // If zero return zero if (n == 0) { return ("Zero"); } // Array to store the powers of 10 String[] multiplier = { "", "Trillion", "Billion", "Million", "Thousand" }; // Array to store numbers till 20 String[] first_twenty = { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }; // Array to store multiples of ten String[] tens = { "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" }; // If number is less than 20, return without any // further computation if (n < 20L) { return (first_twenty[(int)n]); } var answer = ""; for (long i = n; i > 0; i %= limit, limit /= 1000) { // Store the value in multiplier[t], i.e n = // 1000000, then r = 1, for multiplier(million), // 0 for multipliers(trillion and billion) // multiplier here refers to the current // accessible limit curr_hun = i / limit; // It might be possible that the current // multiplier is bigger than your number while (curr_hun == 0) { // Set i as the remainder obtained when n // was divided by the limit i %= limit; // Divide the limit by 1000, shifts the // multiplier limit /= 1000; // Get the current value in hundreds, as // English system works in hundreds curr_hun = i / limit; // Shift the multiplier ++t; } // If current hundred is greater than 99, Add // the hundreds' place if (curr_hun > 99) { answer += (first_twenty[(int)((int)curr_hun / 100)] + " Hundred "); } // Bring the current hundred to tens curr_hun = curr_hun % 100; // If the value in tens belongs to [1,19], add // using the first_twenty if (curr_hun > 0 && curr_hun < 20) { answer += (first_twenty[(int)curr_hun] + " "); } else if (curr_hun % 10 == 0 && curr_hun != 0) { answer += (tens[(int)((int)curr_hun / 10) - 1] + " "); } else if (curr_hun > 20 && curr_hun < 100) { answer += (tens[(int)((int)curr_hun / 10) - 1] + " " + first_twenty[(int)curr_hun % 10] + " "); } // If Multiplier has not become less than 1000, // shift it if (t < 4) { answer += (multiplier[(int)++t] + " "); } } return (answer); } // Driver code public static void Main(String[] args) { // Input 1 var n = 36L; Console.WriteLine(numtowords.numberToWords(n)); // Input 2 n = 123456789; Console.WriteLine(numtowords.numberToWords(n)); // Input 3 n = 10101010110001L; Console.WriteLine(numtowords.numberToWords(n)); // Input 4 n = 999999999; Console.WriteLine(numtowords.numberToWords(n)); }} |
Javascript
function numberToWords(n){ let limit = 1000000000000, t = 0 // If zero console.log zero if (n == 0) { console.log("zero") return } // Array to store the powers of 10 let multiplier = ["", "Trillion", "Billion", "Million", "Thousand"] // Array to store numbers till 20 let first_twenty = ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] // Array to store multiples of ten let tens = ["", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] // If number is less than 20, console.log without any if (n < 20) { console.log(first_twenty[n]) return } let answer = "" let i = n while(i > 0) { /* Store the value in multiplier[t], i.e n = 1000000, then r = 1, for multiplier(million), 0 for multipliers(trillion and billion) multiplier here refers to the current accessible limit */ let curr_hun = Math.floor(i / limit) // It might be possible that the current multiplier is bigger than your number while (curr_hun == 0) { // Set i as the remainder obtained when n was divided by the limit i %= limit // Divide the limit by 1000, shifts the multiplier limit /= 1000 // Get the current value in hundreds, as English system works in hundreds curr_hun = Math.floor(i / limit) // Shift the multiplier t += 1 } let flr = Math.floor(curr_hun / 100); // If current hundred is greater than 99, Add the hundreds' place if (curr_hun > 99) answer += (first_twenty[flr] + " tensundred ") // Bring the current hundred to tens curr_hun = curr_hun % 100 // If the value in tens belongs to [1,19], add using the first_twenty if (curr_hun > 0 && curr_hun < 20) answer += (first_twenty[curr_hun] + " ") // If curr_hun is now a multiple of 10, but not 0 // Add the tens' value using the tens array else if (curr_hun % 10 == 0 && curr_hun != 0){ flr = Math.floor(curr_hun / 10); answer += (tens[flr - 1] + " ") } // If the value belongs to [21,99], excluding the multiples of 10 // Get the ten's place and one's place, and console.log using the first_twenty array else if (curr_hun > 19 && curr_hun < 100){ flr = Math.floor(curr_hun / 10); answer += (tens[flr - 1] + " " + first_twenty[curr_hun % 10] + " ") } // If Multiplier has not become less than 1000, shift it if (t < 4) answer += (multiplier[t] + " ") i = i % limit limit = Math.floor(limit / 1000) } console.log(answer)}// Input 1let n = 36numberToWords(n)n = 123456789// Input 2numberToWords(n)n = 10101010110001// Input 3numberToWords(n)// Input 4n = 999999999numberToWords(n)// This code is contributed by phasing17. |
Output
Thirty Six One Hundred Twenty Three Million Four Hundred Fifty Six Thousand Seven Hundred Eighty Nine Ten Trillion One Hundred One Billion Ten Million One Hundred Ten Thousand One Nine Hundred Nine...
Time Complexity: O(Log10(N))
Auxiliary Space: O(1)
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