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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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     
} 

Python3

# 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 

C#

// 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 

PHP

Output:

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