comments | difficulty | edit_url | rating | source | tags | ||
---|---|---|---|---|---|---|---|
true |
Medium |
1874 |
Weekly Contest 129 Q2 |
|
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
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
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;
}
};
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;
}