Nth number made up of odd digits only

Given an integer N, the task is to find the Nth number made up of odd digits (1, 3, 5, 7, 9) only.
First few numbers made up of odd digits are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, …

Examples:

Input: N = 7
Output: 13
1, 3, 5, 7, 9, 11, 13
13 is the 7th number in the series

Input: N = 10
Output: 19

Approach 1 (Simple) : Starting from 1, keep checking if the number is made up of only odd digits (1, 3, 5, 7, 9) and stop when nth such number is found.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find nth number made up of odd digits only
#include <bits/stdc++.h>
using namespace std;
  
// Function to return nth number made up of odd digits only
int findNthOddDigitNumber(int n)
{
  
    // Variable to keep track of how many
    // such elements have been found
    int count = 0;
    for (int i = 1;; i++) {
        int num = i;
        bool isMadeOfOdd = true;
  
        // Checking each digit of the number
        while (num != 0) {
  
            // If 0, 2, 4, 6 or 8 is found
            // then the number is not made up of odd digits
            if (num % 10 == 0
                || num % 10 == 2
                || num % 10 == 4
                || num % 10 == 6
                || num % 10 == 8) {
                isMadeOfOdd = false;
                break;
            }
  
            num = num / 10;
        }
  
        // If the number is made up of odd digits only
        if (isMadeOfOdd == true)
            count++;
  
        // If it is the nth number
        if (count == n)
            return i;
    }
}
  
// Driver Code
int main()
{
    int n = 10;
    cout << findNthOddDigitNumber(n);
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java program to find nth number  


// made up of odd digits only  


  


import java.io.*; 


  


class GFG { 


    // Function to return nth number made up of odd digits only  


static int findNthOddDigitNumber(int n)  


{  


  


    // Variable to keep track of how many  


    // such elements have been found  


    int count = 0;  


    for (int i = 1;; i++) {  


        int num = i;  


        boolean isMadeOfOdd = true;  


  


        // Checking each digit of the number  


        while (num != 0) {  


  


            // If 0, 2, 4, 6 or 8 is found  


            // then the number is not made up of odd digits  


            if (num % 10 == 0 


                || num % 10 == 2 


                || num % 10 == 4 


                || num % 10 == 6 


                || num % 10 == 8) {  


                isMadeOfOdd = false;  


                break;  


            }  


  


            num = num / 10;  


        }  


  


        // If the number is made up of odd digits only  


        if (isMadeOfOdd == true)  


            count++;  


  


        // If it is the nth number  


        if (count == n)  


            return i;  


    }  


}  


  


// Driver Code  


      


    public static void main (String[] args) { 


    int n = 10;  


    System.out.println (findNthOddDigitNumber(n));  


          


    } 


//This code is contributed by ajit     


} 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python 3 program to find nth number  


# made up of odd digits only 


  


# Function to return nth number made 


# up of odd digits only 


def findNthOddDigitNumber(n) : 


      


    # Variable to keep track of how many 


    # such elements have been found 


    count = 0


      


    i = 1


    while True : 


        num = i 


        isMadeOfOdd = True


          


        # Checking each digit of the number 


        while num != 0 : 


              


            # If 0, 2, 4, 6 or 8 is found 


            # then the number is not made 


            # up of odd digits 


            if (num % 10 == 0 or num % 10 == 2 or 


                num % 10 == 4 or num % 10 == 6 or


                num % 10 == 8) : 


                      


                    isMadeOfOdd = False


                    break


          


            num /= 10


      


        # If the number is made up of 


        # odd digits only 


        if isMadeOfOdd == True : 


            count += 1


      


        # If it is the nth number 


        if count == n : 


            return i 


      


        i += 1


  


# Driver code 


if __name__ == "__main__" : 


      


    n = 10


      


    # Function call 


    print(findNthOddDigitNumber(n)) 


      


# This code is contributed by Ryuga 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# program to find nth number  


// made up of odd digits only  


using System; 


  


class GFG 


{ 


      


// Function to return nth number  


// made up of odd digits only  


static int findNthOddDigitNumber(int n)  


{  


  


    // Variable to keep track of  


    // how many such elements have 


    // been found  


    int count = 0;  


    for (int i = 1;; i++) 


    {  


        int num = i;  


        bool isMadeOfOdd = true;  


  


        // Checking each digit  


        // of the number  


        while (num != 0)  


        {  


  


            // If 0, 2, 4, 6 or 8 is found  


            // then the number is not made  


            // up of odd digits  


            if (num % 10 == 0 || num % 10 == 2 ||  


                num % 10 == 4 || num % 10 == 6 ||  


                num % 10 == 8) 


            {  


                isMadeOfOdd = false;  


                break;  


            }  


  


            num = num / 10;  


        }  


  


        // If the number is made up of 


        // odd digits only  


        if (isMadeOfOdd == true)  


            count++;  


  


        // If it is the nth number  


        if (count == n)  


            return i;  


    }  


}  


  


// Driver Code  


static public void Main () 


{ 


    int n = 10;  


    Console.WriteLine(findNthOddDigitNumber(n));  


} 


} 


  


// This code is contributed  


// by Ajit Deshpal 



















                        chevron_right







                        filter_none









PHP










Output:


19





Approach 2 (Queue Based):  The idea is to generate all numbers (smaller than n) containing odd digits only. How to generate all numbers smaller than n with odd digits? We use queue for this. Initially we push ‘1’, ‘3’, ‘5’, ‘7’ and ‘9’ to the queue. Then we run a loop while count of processed elements is smaller than n. We pop an item one by one and for every popped item x, we generate next numbers x*10 + 1, x*10 + 3, x*10 + 5, x*10 + 7 and x*10 + 9. We enqueue these new numbers. Time complexity of this approach is O(n)


Please refer below post for implementation of this approach.

Count of Binary Digit numbers smaller than N



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Shivam.Pradhan
Discovering ways to develop a plane for soaring career goals
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.