<p>Given an integer <code>n</code>, return <em>the number of positive integers in the range </em><code>[1, n]</code><em> that have <strong>at least one</strong> repeated digit</em>.</p>

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

<pre>
<strong>Input:</strong> n = 20
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only positive number (&lt;= 20) with at least 1 repeated digit is 11.
</pre>

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

<pre>
<strong>Input:</strong> n = 100
<strong>Output:</strong> 10
<strong>Explanation:</strong> The positive numbers (&lt;= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
</pre>

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

<pre>
<strong>Input:</strong> n = 1000
<strong>Output:</strong> 262
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def dfs(self, s: str, mask: int, idx: int, is_num: bool, is_limit: bool) -> int:
        if idx == len(s):
            return 1 if is_num else 0
        elif (mask, idx, is_num, is_limit) in self.cache:
            return self.cache[(mask, idx, is_num, is_limit)]
        else:
            total = 0
            if not is_num:
                total += self.dfs(s, mask, idx+1, False, False)
            min_digit = 1 if not is_num else 0
            max_digit = int(s[idx]) if is_limit else 9
            for digit in range(min_digit, max_digit+1):
                if (mask & (1<<digit)) != 0:
                    continue
                total += self.dfs(s, (mask | (1<<digit)), idx+1, True, is_limit and digit==max_digit)
            self.cache[(mask, idx, is_num, is_limit)] = total
            return total
    
    def numDupDigitsAtMostN(self, n: int) -> int:
        self.cache = {}
        s = str(n)
        res = self.dfs(s, 0, 0, False, True)
        return n-res
17100