<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:
        digits = list(map(int, str(n)))
        length = len(digits)
        memo = {}

        def dp(pos, mask, tight):
            if pos == length:
                return 1  # Reached the end, valid number

            if (pos, mask, tight) in memo:
                return memo[(pos, mask, tight)]

            limit = digits[pos] if tight else 9
            result = 0

            for digit in range(0, limit + 1):
                if mask == 0 and digit == 0:
                    # Leading zero case
                    result += dp(pos + 1, mask, tight and (digit == limit))
                elif mask & (1 << digit) == 0:
                    # Use this digit if it hasn't been used before
                    result += dp(pos + 1, mask | (1 << digit), tight and (digit == limit))

            memo[(pos, mask, tight)] = result
            return result

        return n - dp(0, 0, True) + 1
20300