<p>We call a positive integer <strong>special</strong> if all of its digits are <strong>distinct</strong>.</p>

<p>Given a <strong>positive</strong> integer <code>n</code>, return <em>the number of special integers that belong to the interval </em><code>[1, n]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 20
<strong>Output:</strong> 19
<strong>Explanation:</strong> All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong> All the integers from 1 to 5 are special.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 135
<strong>Output:</strong> 110
<strong>Explanation:</strong> There are 110 integers from 1 to 135 that are special.
Some of the integers that are not special are: 22, 114, and 131.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>9</sup></code></li>
</ul>

class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        num = list(map(int, str(n)))

        @cache
        def dp(pos: int, tight: bool, chosen_digit: int) -> int:
            if pos == len(num): return 1
            res = 0
            bound = num[pos] if tight else 10
            for i in range(bound + 1):
                if i == 10: continue
                if chosen_digit & (1 << i): continue 
                new_tight = i == bound
                new_chosen_digit = 0 if chosen_digit == 0 and i == 0 \
                                     else chosen_digit | (1 << i)
                res += dp(pos + 1, new_tight, new_chosen_digit)
            return res

        return dp(0, True, 0) - 1
242