Given a positive number n. We need to find x such that 1*n, 2*n, 3*n…..x*n gives all 10 digits at least once. If no such x is possible print -1.
Input : n = 1692 Output : 3 Explanation: n = 1692, we got the digits- 1, 2, 6, 9 2*n = 3384, we got the digits- 1, 2, 3, 4, 6, 8, 9. 3*n = 5076, we got the digits- 1, 2, 3, 4, 5, 6, 7, 8, 9. At this step we got all the digits at least once. Therefore our answer is 3. Input : 1 Output : 10 Input : 0 Output :-1
The idea used here is simple. We start from 1 and keep multiplying with n till we do not get all the 10 digits at least once. In order to keep track of all the digits coming at each iteration we use an temporary array of size 10 initially having all zeroes. Whenever we got a digit first time we will initialize its index in array with 1. When all digits are visited once, we are done.
Following is the implementation of it.
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