smallest_integer_divisibe_k.md

1015. Smallest Integer Divisible by K

• let the number be n.
• to keep 1 in no. keep adding multiplying like n = n * 10 + 1.
• to avoid memory limit to be exceeded keep modulating n with k.
• if we encounter any multiple of k we immediately will get 0.

Code implementation

class Solution {
public:
    int smallestRepunitDivByK(int k) {
        if (k % 2 == 0 or k % 5 == 0) return -1;

        long long int num = 1;
        int ans = 1;

        if (num % k == 0) return ans;

        while (num) {
            num = num * 10 + 1;
            num %= k;
            ans ++;
        }
        return ans;
    }
};