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 seriesInput: 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; } |
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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 } |
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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 |
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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 |
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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
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