<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 numDupDigitsAtMostN(self, n: int) -> int:
        num = str(n)
        @cache
        def dp(indx, mask, tight):
            if indx == len(num):
                return 1
            upper_bound = int(num[indx]) if tight else 9
            ans = 0
            for digit in range(upper_bound + 1):
                if mask & (1 << digit): continue
                new_mask = mask if (mask == 0 and digit == 0) else mask | (1 << digit)
                ans += dp(indx + 1, new_mask, tight and (digit == upper_bound))
            return ans
        return n + 1 - dp(0, 0, True)
21100