0ms C++ SOLUTION
Intuition\n Describe your first thoughts on how to solve this problem. \nThe idea of this solution is to count the number of integers with unique digits from 0 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe idea of this solution is to count the number of integers with unique digits from 0 to N, which can be done by considering the different cases for each digit. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe start by looping through the digits of N, starting from the least significant digit. For each digit, we first calculate the number of integers with unique digits that are smaller than N and have the same number of digits as N. For this, we use a helper function `A(m, n)` which calculates the number of numbers with digits 1, 2, ..., m, where each digit can appear at most once and the total number of digits is n. \n\nThen, for the current digit, we loop through all possible digits (skipping 0 if the current digit is the most significant digit) and check if it has already been seen. If yes, we break, as it means that all the remaining digits should be the same as the current digit. Otherwise, we increment the count by the number of integers with unique digits that are smaller than N and have the same number of digits as N. \n\nFinally, we subtract the count from N to get the result.\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int numDupDigitsAtMostN(int n) {\n        vector<int> digits;\n        for (int x = n + 1; x > 0; x /= 10) {\n            digits.push_back(x % 10);\n        }\n        reverse(digits.begin(), digits.end());\n        int ans = 0, k = digits.size();\n        for (int i = 1; i < k; ++i) {\n            ans += 9 * A(9, i - 1);\n        }\n        set<int> seen;\n        for (int i = 0; i < k; ++i) {\n            for (int j = i > 0 ? 0 : 1; j < digits[i]; ++j) {\n                if (!seen.count(j)) {\n                    ans += A(9 - i, k - i - 1);\n                }\n            }\n            if (seen.count(digits[i])) {\n                break;\n            }\n            seen.insert(digits[i]);\n        }\n        return n - ans;\n    }\n    int A(int m, int n) {\n        return n == 0 ? 1 : A(m, n - 1) * (m - n + 1);\n    }\n};\n```
['C++']