# Question

Formatted question description: https://leetcode.ca/all/357.html

 357	Count Numbers with Unique Digits

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

Example:

Input: 2
Output: 91
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100,
excluding 11,22,33,44,55,66,77,88,99

@tag-dp
@tag-math


# Algorithm

Three digits

11 can be 110 or 101
22 can be 220 or 202
So there are only 8 possibilities

10 +
90-9 +
(90-9) * 8


In four digits

111 can be 1010, 1101, 1110


There are only 7 possibilities

• One-digit number that meets the requirements is 10 (0 to 9)
• The two-digit number that satisfies the meaning of the question is 81, [10-99] remove the [11,22,33,44,55,66,77,88,99] of these 90 numbers, and there is still 81.

The general term formula is f(k) = 9 * 9 * 8 * ... (9-k + 2), then we can pass the [1, n] interval digits through the general term formula according to the size of n Calculate and add up

# Code

Java

• 
public class Count_Numbers_with_Unique_Digits {

class Solution {

public int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
int res = 0;
for (int i = 1; i <= n; ++i) {
res += count(i);
}
return res;
}

int count(int k) {
if (k < 1) {
return 0;
}
if (k == 1) {
return 10;
}

int res = 1;
for (int i = 9; i >= (11 - k); --i) {
res *= i;
}
return res * 9;
}
}
}

• Todo

• class Solution:
def countNumbersWithUniqueDigits(self, n: int) -> int:
if n == 0:
return 1
if n == 1:
return 10
ans, cur = 10, 9
for i in range(n - 1):
cur *= 9 - i
ans += cur
return ans

############

class Solution(object):
def countNumbersWithUniqueDigits(self, n):
"""
:type n: int
:rtype: int4
"""
if n <= 1:
return 10 ** n
dp =  * (n + 1)
dp = 0
dp = 9
k = 9
for i in range(2, n + 1):
dp[i] = max(dp[i - 1] * k, 0)
k -= 1
return sum(dp) + 1