<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:
        def count_special(n):
            s = str(n)
            @cache
            def dp(i, mask, is_limit, is_num):
                if i == len(s):
                    return is_num
                res = 0
                if not is_num:
                    res = dp(i + 1, mask, False, False)
                low = 0 if is_num else 1
                up = int(s[i]) if is_limit else 9
                for d in range(low, up + 1):
                    if (mask >> d) & 1 == 0:
                        res += dp(i + 1, mask | (1 << d), is_limit and d == up, True)
                return res
            return dp(0, 0, True, False)
        return n - count_special(n)
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