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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;
}

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