# 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:**

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Input:N = 7Output:13

1, 3, 5, 7, 9, 11, 13

13 is the 7th number in the series

Input:N = 10Output: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