Formatted question description: https://leetcode.ca/all/1015.html

1015. Smallest Integer Divisible by K

Level

Medium

Description

Given a positive integer K, you need find 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.

Example 1:

Input: 1

Output: 1

Explanation: The smallest answer is N = 1, which has length 1.

Example 2:

Input: 2

Output: -1

Explanation: There is no such positive integer N divisible by 2.

Example 3:

Input: 3

Output: 3

Explanation: The smallest answer is N = 111, which has length 3.

Note:

  • 1 <= K <= 10^5

Solution

If K is a multiple of 2 or 5, then any number divisible by K can’t end with 1, so return -1.

Otherwise, each time append a 1 to the current number (which is initially 0) and check whether the new number is divisible by K. Use modulo operation to avoid overflow.

  • class Solution {
        public int smallestRepunitDivByK(int K) {
            if (K % 2 == 0 || K % 5 == 0)
                return -1;
            int value = 1;
            int length = 1;
            while (value % K != 0) {
                value = value % K * 10 + 1;
                length++;
            }
            return length;
        }
    }
    
  • // OJ: https://leetcode.com/problems/smallest-integer-divisible-by-k/
    // Time: O(K)
    // Space: O(K)
    class Solution {
    public:
        int smallestRepunitDivByK(int k) {
            unordered_set<int> s;
            for (int n = 1 % k, len = 1; true; n = (n * 10 + 1) % k, ++len) {
                if (n == 0) return len;
                if (s.count(n)) return -1;
                s.insert(n);
            }
        }
    };
    
  • # 1015. Smallest Integer Divisible by K
    # https://leetcode.com/problems/smallest-integer-divisible-by-k/
    
    class Solution:
        def smallestRepunitDivByK(self, k: int) -> int:
            if k % 2 == 0 and k % 5 == 0: return -1
            mod_set = set()
            prev = 0
            
            for i in range(1, k + 1):
                prev = (prev * 10 + 1) % k
                if prev == 0: return i
                if prev in mod_set: return -1
                mod_set.add(prev)
            
            return -1
    
    

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