<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:
        digits = []
        while n > 0:
            digits.append(n % 10)
            n //= 10
        digits.reverse()
        
        @cache
        def dp(i, tight, leadingZero, usedMask):
            if i == len(digits):
                return 1
            
            maxDigits = digits[i] if tight else 9
            ans = 0
            for d in range(maxDigits + 1):
                if (usedMask >> d) & 1:
                    continue
                nextTight = tight and d == maxDigits
                nextLeadingZero = leadingZero and d == 0
                newUsedMask = usedMask
                if not nextLeadingZero:
                    newUsedMask = usedMask | (1 << d)
                ans += dp(i + 1, nextTight, nextLeadingZero, newUsedMask)
            return ans
        return dp(0, True, True, 0) - 1
230