Question

1015. Smallest Integer Divisible by K

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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

Solution 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

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;
}