Given a number N. The task is to find the smallest N digit ODD number.
Input: N = 1 Output: 1 Input: N = 3 Output: 101
Approach: There can be two cases depending on the value of N.
Case 1 : If N = 1 then answer will be 1.
Case 2 : If N != 1 then answer will be (10^(n-1)) + 1 because the series of the smallest odd numbers will go on like: 1, 11, 101, 1001, 10001, 100001, ….
Below is the implementation of the above approach:
Time Complexity: O(1).
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