<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 f2(n):
            a = list(map(int, str(n)))[::-1]
            l = len(a)
            @cache
            def f(i, s, g, m, z):
                if i < 0:
                    return +g
                ans = 0
                for d in range(10):
                    if not z and d == 0 or m and d > a[i]:
                        continue
                    if m and d == a[i]:
                        ans += f(i - 1, s | 1 << d, g or (s >> d & 1) != 0, True, True)
                    else:
                        ans += f(i - 1, s | 1 << d, g or (s >> d & 1) != 0, False, True)
                return ans
            return f(l - 1, 0, False, True, False)
        ans = f2(n)
        k = 9
        while k < n:
            ans += f2(k)
            k = k * 10 + 9
        return ans
27000