Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 25

Description

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

Solution

class Solution {
public:
  int smallestRepunitDivByK(int k) {
    if(k % 2 == 0 || k % 5 == 0) return -1;
    int cur = 0;
    for(int i = 1; i <= k; ++i) {
      cur = (cur * 10 + 1) % k;
      if(cur == 0) return i;
    }
    // actually won't go to here
    return INT_MAX;
  }
};

// Accepted
// 70/70 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 46.22 % of cpp submissions (6 MB)