Given an integer N, the task is to check if the concatenation of first N natural numbers is divisible by 3 or not. Print Yes if divisible and No if not.
Input: N = 3
The concatenated number = 123
Since it is divisible by 3, the output is Yes
Input: N = 7
Explanation: The concatenated number = 1234567
Since it is not divisible by 3, the output is No.
The simplest approach is to concatenate the first N natural numbers and calculate the sum of digits of the resultant number and check if it is divisible by 3 or not.
Time Complexity: O(N)
Auxiliary Space: O(1)
To optimize the above approach, we can observe a pattern. The concatenation of first N natural numbers is not divisible by 3 for the following series 1, 4, 7, 10, 13, 16, 19, and so on. The Nth term of the series is given by the formula 3×n +1. Hence, if (N – 1) is not divisible by 3, then the resultant number is divisible by 3, so print Yes. Otherwise, print No.
Below is the implementation of the above approach:
Time Complexity: O(1)
Auxiliary Space: O(1)
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