Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10ⁿ.

Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])

Solution: Math

f(0) = 1 (0)

f(1) = 10 (0 – 9)

f(2) = 9 * 9 (1-9 * (0 ~ 9 exclude the one from first digit))

f(3) = 9 * 9 * 8

f(4) = 9 * 9 * 8 * 7

…

f(x) = 0 if x >= 10

ans = sum(f[1] ~ f[n])

Time complexity: O(1)

Space complexity: O(1)

// Author: Huahua
// Running time: 0 ms
class Solution {
public:
  int countNumbersWithUniqueDigits(int n) {
    if (n == 0) return 1;
    vector<int> f(11);    
    f[1] = 10;
    f[2] = 9 * 9;
    for (int i = 3; i <= 10; ++i)
      f[i] = f[i - 1] * (10 - i + 1);
    int ans = 0;
    for (int i = 0; i <= min(10, n); ++i)
      ans += f[i];
    return ans;
  }
};

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