smallest-greater-multiple-made-of-two-digits.cpp

// Time:  sum(O(l * 2^l) for l in range(1, 11)) = O(20 * 2^10) = O(1)
// Space: O(1)

class Solution {
public:
    int findInteger(int k, int digit1, int digit2) {
        static const int MAX_NUM_OF_DIGITS = 10;

        if (digit1 < digit2) {
            swap(digit1, digit2);
        }
        for (int l = 1, total = 2; l <= MAX_NUM_OF_DIGITS; ++l, total <<= 1) {
            for (int mask = 0; mask < total; ++mask) {
                int64_t curr = 0;
                for (int bit = total >> 1; bit; bit >>= 1) {
                    curr = curr * 10 + ((mask & bit) ? digit1 : digit2);
                }
                if (k < curr && curr <= numeric_limits<int>::max() && curr % k == 0) {
                    return curr;
                }
            }
        }
        return -1; 
    }
};