<p>Given an integer <code>n</code>, return <em>the number of positive integers in the range </em><code>[1, n]</code><em> that have <strong>at least one</strong> repeated digit</em>.</p>

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

<pre>
<strong>Input:</strong> n = 20
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only positive number (&lt;= 20) with at least 1 repeated digit is 11.
</pre>

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

<pre>
<strong>Input:</strong> n = 100
<strong>Output:</strong> 10
<strong>Explanation:</strong> The positive numbers (&lt;= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
</pre>

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

<pre>
<strong>Input:</strong> n = 1000
<strong>Output:</strong> 262
</pre>

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

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

class Solution {
public:
    int dp[11][2][1025]; // 1025 -> (1 << 10)
    int dfs(int i, int mask, bool tight, string& digits) {
        if (i == digits.size()) 
        {
            return mask == 0 ? 0 : 1;
        }
        if (dp[i][tight][mask] != -1)
            return dp[i][tight][mask];
        int limit = 9, res = 0;
        if (tight)
            limit = (digits[i] - '0');
        for (int dig = 0; dig <= limit; dig++) 
        {
            if (mask == 0 && dig == 0) 
            {
                res += dfs(i + 1, mask, (tight & (limit == dig)), digits);
                continue;
            }
            int repeated = mask & (1 << dig);
            if (repeated == 0) 
            {
                res += dfs(i + 1, mask | (1 << dig), (tight & (limit == dig)), digits);
            }
        }
        return dp[i][tight][mask] = res;
    }
    int numDupDigitsAtMostN(int n) {
        memset(dp, -1, sizeof dp);
        string x = to_string(n);
        return n - dfs(0, 0, true, x);
    }
};
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