Count Numbers with Unique Digits
<p>Given an integer <code>n</code>, return the count of all numbers with unique digits, <code>x</code>, where <code>0 &lt;= x &lt; 10<sup>n</sup></code>.</p>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 91
<strong>Explanation:</strong> The answer should be the total numbers in the range of 0 &le; x &lt; 100, excluding 11,22,33,44,55,66,77,88,99
</pre>

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

<pre>
<strong>Input:</strong> n = 0
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>0 &lt;= n &lt;= 8</code></li>
</ul>
Math; Dynamic Programming; Backtracking
class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        @cache
        def dfs(i: int, mask: int, lead: bool) -> int:
            if i < 0:
                return 1
            ans = 0
            for j in range(10):
                if mask >> j & 1:
                    continue
                if lead and j == 0:
                    ans += dfs(i - 1, mask, True)
                else:
                    ans += dfs(i - 1, mask | 1 << j, False)
            return ans

        return dfs(n - 1, 0, True)
