<p>Given a positive integer <code>k</code>, you need to find the <strong>length</strong> of the <strong>smallest</strong> positive integer <code>n</code> such that <code>n</code> is divisible by <code>k</code>, and <code>n</code> only contains the digit <code>1</code>.</p>

<p>Return <em>the <strong>length</strong> of </em><code>n</code>. If there is no such <code>n</code>, return -1.</p>

<p><strong>Note:</strong> <code>n</code> may not fit in a 64-bit signed integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> The smallest answer is n = 1, which has length 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 2
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no such positive integer n divisible by 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> The smallest answer is n = 111, which has length 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int smallestRepunitDivByK(int k) {
      if(k % 10 != 1 && k % 10 != 3 && k % 10 != 7 && k % 10 != 9)
        return -1;

      unordered_set<int>seen;
      int n=0;

      for(int length=1;length<=k;++length){
        n = (n*10+1)%k;
        if(n==0)
          return length;
        if(seen.count(n))
          return -1;
        seen.insert(n);    
      } 
      return -1;    
    }
};
16100