Algorithm
Problem Name: 1015. Smallest Integer Divisible by K
Given a positive integer k, you need to find the length of the smallest positive integer n such that n is divisible by k, and n only contains the digit 1.
Return the length of n. If there is no such n, return -1.
Note: n may not fit in a 64-bit signed integer.
Example 1:
Input: k = 1 Output: 1 Explanation: The smallest answer is n = 1, which has length 1.
Example 2:
Input: k = 2 Output: -1 Explanation: There is no such positive integer n divisible by 2.
Example 3:
Input: k = 3 Output: 3 Explanation: The smallest answer is n = 111, which has length 3.
Constraints:
1 <= k <= 105
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int smallestRepunitDivByK(int K) {
if (K % 2 == 0 || K % 5 == 0) {
return -1;
}
int remainder = 0;
for (int n = 1; n < = K; n++) {
remainder = (remainder * 10 + 1) % K;
if (remainder == 0) {
return n;
}
}
return -1;
}
}
Input
k = 1
Output
1
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const smallestRepunitDivByK = function(K) {
if (K % 2 === 0 || K % 5 === 0) return -1;
let r = 0;
for (let N = 1; N < = K; ++N) {
r = (r * 10 + 1) % K;
if (r == 0) return N;
}
return -1;
};
Input
k = 1
Output
1
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def smallestRepunitDivByK(self, K: int) -> int:
used, mod, cnt = set(), 1 % K, 1
while mod:
mod = (mod * 10 + 1) % K
cnt += 1
if mod in used:
return -1
used.add(mod)
return cnt
Input
k = 3
Output
3