Welcome to Subscribe On Youtube
1015. Smallest Integer Divisible by K
Description
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
Solutions
-
class Solution { public int smallestRepunitDivByK(int k) { int n = 1 % k; for (int i = 1; i <= k; ++i) { if (n == 0) { return i; } n = (n * 10 + 1) % k; } return -1; } } -
class Solution { public: int smallestRepunitDivByK(int k) { int n = 1 % k; for (int i = 1; i <= k; ++i) { if (n == 0) { return i; } n = (n * 10 + 1) % k; } return -1; } }; -
class Solution: def smallestRepunitDivByK(self, k: int) -> int: n = 1 % k for i in range(1, k + 1): if n == 0: return i n = (n * 10 + 1) % k return -1 -
func smallestRepunitDivByK(k int) int { n := 1 % k for i := 1; i <= k; i++ { if n == 0 { return i } n = (n*10 + 1) % k } return -1 } -
function smallestRepunitDivByK(k: number): number { let n = 1 % k; for (let i = 1; i <= k; ++i) { if (n === 0) { return i; } n = (n * 10 + 1) % k; } return -1; }