Given three integers N, A and B, the task is to find whether N is divisible by any number that contains only A and B as it’s digits.
Input: N = 106, a = 3, b = 5
106 is divisible by 53
Input: N = 107, a = 3, b = 5
Approach 1 (Recursive): An efficient solution is to make all the numbers that contains a and b as their digits using recursive function starting with the numbers a and b. If function call is fun(x) then recursively call for fun(x * 10 + a) and fun(x * 10 + b). If n is divisible by any of the number then print Yes else print No.
Below is the implementation of the above approach:
Approach 2 (Queue Based): The idea is to generate all numbers (smaller than n) containing digits a and b. For every number check if it divides n or not. How to generate all numbers smaller than n? We use queue for this. Initially we push ‘a’ and ‘b’ to the queue. Then we run a loop while front of queue is smaller than n. We pop an item one by one and for ever popped item x, we generate next numbers x*10 + a and x*10 + b and enqueue them. 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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- Number of integers in a range [L, R] which are divisible by exactly K of it's digits
- Possible to make a divisible by 3 number using all digits in an array
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