<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 f(string &n, string s, int i, int isTight, int mask, vector<vector<vector<int>>> &dp) {
        if(i == n.length()) {
            return 1;
        }

        if(dp[i][mask][isTight] != -1) {
            return dp[i][mask][isTight];
        }

        int ub = isTight ? n[i] - '0' : 9;
        int ans = 0;
        for(int dig=0 ; dig<=ub ; dig++) {
            if((s.length() == 0 || stoi(s) == 0) && dig == 0) {
                ans += f(n, s + (char)(dig + '0'), i+1, isTight && (dig == n[i] - '0'), mask, dp);
            }
            else if((mask & (1 << dig)) == 0) {
                ans += f(n, s + (char)(dig + '0'), i+1, isTight && (dig == n[i] - '0'), mask | (1 << dig), dp);
            }
        }

        return dp[i][mask][isTight] = ans;
    }
    int numDupDigitsAtMostN(int n) {
        string num = to_string(n);

        vector<vector<vector<int>>> dp(num.length(), vector<vector<int>>(1024, vector<int>(2, -1)));

        return n - f(num, "", 0, 1, 0, dp) + 1;
    }
};
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