<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:
        
        @lru_cache(1000)
        def dp(s, index, tight, leading_zero, mask = 0):
            if index >= len(s): return int(leading_zero == False)
            bound, res = int(s[index]) if tight else 9, 0

            for digit in range(bound+1):
                if (1 << digit) & mask > 0: continue

                new_tight = tight and int(s[index]) == digit
                new_leading_zero = leading_zero and digit == 0

                if not new_leading_zero: new_mask = mask | (1 << digit)
                else:  new_mask = mask

                res += dp(s, index + 1, new_tight, new_leading_zero, new_mask)

            return res 
                
        return n - dp(str(n), 0, True, True, 0)
19900