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 {
    private Integer[][] f;

    public int countNumbersWithUniqueDigits(int n) {
        f = new Integer[n][1 << 10];
        return dfs(n - 1, 0, true);
    }

    private int dfs(int i, int mask, boolean lead) {
        if (i < 0) {
            return 1;
        }
        if (!lead && f[i][mask] != null) {
            return f[i][mask];
        }
        int ans = 0;
        for (int j = 0; j <= 9; ++j) {
            if ((mask >> j & 1) == 1) {
                continue;
            }
            if (lead && j == 0) {
                ans += dfs(i - 1, mask, true);
            } else {
                ans += dfs(i - 1, mask | 1 << j, false);
            }
        }
        if (!lead) {
            f[i][mask] = ans;
        }
        return ans;
    }
}
