Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

Nth number made up of odd digits only

  • Difficulty Level : Medium
  • Last Updated : 31 Mar, 2021

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++




// 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



# Python3 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



<?php
// PHP program to find nth number
// made up of odd digits only
 
// Function to return nth number
// made up of odd digits only
function 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;
            }
            $num = (int)($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
$n = 10;
 
// Function call
print(findNthOddDigitNumber($n));
     
// This code is contributed by mits
?>
Javascript



<script>
 
// JavaScript program to find nth
// number made up of odd digits only
 
// Function to return nth number
// made up of odd digits only
function findNthOddDigitNumber(n)
{
     
    // Variable to keep track of how many
    // such elements have been found
    let count = 0;
    for(let i = 1;; i++)
    {
        let num = i;
        let 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 = Math.floor(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
let n = 10;
 
document.write(findNthOddDigitNumber(n));
 
// This code is contributed by Manoj.
 
</script>
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
 
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes
arrow_drop_up
Next

Finding n-th number made of prime digits (2, 3, 5 and 7) only
Recommended Articles
Page :
Number of digits in the nth number made of given four digits
07, May 18
Minimum number closest to N made up of odd digits only
26, Aug 20
Count N-digits numbers made up of even and prime digits at odd and even positions respectively
29, Jul 21
Count of integers in a range which have even number of odd digits and odd number of even digits
21, Aug 19
Count N-digit numbers made up of X or Y whose sum of digits is also made up of X or Y
06, Nov 20
Find the n-th number made of even digits only
27, May 17
Finding n-th number made of prime digits (2, 3, 5 and 7) only
24, Jun 17
Smallest multiple of a given number made of digits 0 and 9 only
21, Mar 18
Numbers of Length N having digits A and B and whose sum of digits contain only digits A and B
31, Jan 19
Count numbers from given range having odd digits at odd places and even digits at even places
02, Dec 20
Smallest positive number made up of non-repeating digits whose sum of digits is N
17, Dec 20
Largest element in the longest Subarray consisting of only Even or only Odd numbers
09, Dec 20
Number formed by deleting digits such that sum of the digits becomes even and the number odd
26, Feb 20
Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
13, Feb 19
Sum of the digits of square of the given number which has only 1's as its digits
20, Sep 19
Count N-length arrays made from first M natural numbers whose subarrays can be made palindromic by replacing less than half of its elements
16, Dec 20
Generate an N-digit number made up of 1 or 2 only which is divisible by 2N
09, Feb 21
Check whether product of digits at even places is divisible by sum of digits at odd place of a number
19, Nov 18
Sum of nth terms of Modified Fibonacci series made by every pair of two arrays
30, Oct 18
Make all the elements of array odd by incrementing odd-indexed elements of odd-length subarrays
16, Dec 20
Count of Rectangles with area K made up of only 1s from given Binary Arrays
27, Aug 20
Count subarrays made up of single-digit integers only
01, Feb 21
Count numbers in given range such that sum of even digits is greater than sum of odd digits
26, Jun 19
Maximum number made up of distinct digits whose sum is equal to N
25, May 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
Shivam.Pradhan
@Shivam.Pradhan
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!