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, …
Input: N = 7
1, 3, 5, 7, 9, 11, 13
13 is the 7th number in the series
Input: N = 10
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:
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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