LeetCode 357. Count Numbers with Unique Digits

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])

Seems like there is a rule for it.

if n == 1, there will be 10 numbers [0 -- 9]

if n == 2, since we can't start from 0, if we start from 1, there will be 1*9, start with 2, there will be 1 * 9..... thus, n == 2, in total = 1*9*9

if n ==3, since we can't start from 0, if we start from 1*, there will be 1*9*8, start with 2, there will be 1*9*8 ... thus n == 3, in total = 1*9*9*8

if n == 4, since we can't start from 0, if start with 1**, there will be 1*9*8*7, start with 2, 1*9*8*7.... n ==3, in total = 1*9*9*8*7

.... the total is n==1 + n==2 + ... n == n

int fn(n) {
  int n = 2;
  int prev = 10;
  int sum = 9;
 while(n > 1) {
    sum = sum * (9 - n + 2);
    total = prev + sum;
    n = n - 1;
  }
  return total;
}

Or use dp.

    int countNumbersWithUniqueDigits(int n) {
        if(!n) return 1;
        vector<int> dp(n, 0);
        dp[0] = 9;
        for(int i = 1; i < n; ++i) {
            dp[i] = dp[i-1] * (10 - i);
        }
        int sum = 0;
        for(int i = 0; i < n; ++i) sum += dp[i];
        return sum + 1;
    }