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 {
public:
    int countNumbersWithUniqueDigits(int n) {
        int f[n + 1][1 << 10];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i, int mask, bool lead) -> int {
            if (i < 0) {
                return 1;
            }
            if (!lead && f[i][mask] != -1) {
                return f[i][mask];
            }
            int ans = 0;
            for (int j = 0; j <= 9; ++j) {
                if (mask >> j & 1) {
                    continue;
                }
                if (lead && j == 0) {
                    ans += dfs(i - 1, mask, true);
                } else {
                    ans += dfs(i - 1, mask | 1 << i, false);
                }
            }
            if (!lead) {
                f[i][mask] = ans;
            }
            return ans;
        };
        return dfs(n - 1, 0, true);
    }
};
