# 357. Count Numbers with Unique Digits

## Description

Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n.

Example 1:

Input: n = 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


Example 2:

Input: n = 0
Output: 1


Constraints:

• 0 <= n <= 8

## Solutions

• class Solution {
public int countNumbersWithUniqueDigits(int n) {
if (n == 0) {
return 1;
}
if (n == 1) {
return 10;
}
int ans = 10;
for (int i = 0, cur = 9; i < n - 1; ++i) {
cur *= (9 - i);
ans += cur;
}
return ans;
}
}

• class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
if (n == 1) return 10;
int ans = 10;
for (int i = 0, cur = 9; i < n - 1; ++i) {
cur *= (9 - i);
ans += cur;
}
return ans;
}
};

• 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


• func countNumbersWithUniqueDigits(n int) int {
if n == 0 {
return 1
}
if n == 1 {
return 10
}
ans := 10
for i, cur := 0, 9; i < n-1; i++ {
cur *= (9 - i)
ans += cur
}
return ans
}