LeetCode 1015: Smallest Integer Divisible by K Solution

Master LeetCode problem 1015 (Smallest Integer Divisible by K), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1015. Smallest Integer Divisible by K

Medium

Hash Table Math

Problem Explanation

Explanation:

To solve this problem, we can use the idea of remainders. We iterate through all possible values of n until we find a number that is divisible by k. During each iteration, we calculate the remainder of the current number when divided by k. If the remainder is 0, we return the length of the current number. If we encounter a number that we have already seen before, it means we are in a loop and there is no valid answer.

Algorithm:

Initialize a remainder variable to 0.
Iterate from 1 to k.
Calculate the new remainder by multiplying the previous remainder by 10, adding 1, and then taking the modulo k.
If the new remainder is 0, return the current iteration count.
If we encounter a remainder that we have seen before, return -1.

Time Complexity: O(k)
Space Complexity: O(1)

Solution Code

class Solution {
    public int smallestRepunitDivByK(int k) {
        int remainder = 0;
        for (int i = 1; i <= k; i++) {
            remainder = (remainder * 10 + 1) % k;
            if (remainder == 0) {
                return i;
            }
        }
        return -1;
    }
}

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