<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:
        dp = [[[-1] * 1025 for _ in range(2)] for _ in range(11)] # 1025 -> (1 << 10)

        def dfs(i: int, mask: int, tight: bool, digits: str) -> int:
            if i == len(digits):
                return 0 if mask == 0 else 1
            
            if dp[i][tight][mask] != -1:
                return dp[i][tight][mask]

            limit, res = 9, 0
            if tight:
                limit = int(digits[i])  # '0' - '0' -> int('0')
            
            for dig in range(limit + 1):
                if mask == 0 and dig == 0:
                    res += dfs(i + 1, mask, (tight and (limit == dig)), digits)
                    continue
            
                repeated = mask & (1 << dig)
                if repeated == 0:
                    res += dfs(i + 1, mask | (1 << dig), (tight and (limit == dig)), digits)
            dp[i][tight][mask] = res
            return res

        x = str(n)
        return n - dfs(0, 0, True, x)
21600