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:
Solutions#
Solution 1#
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
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Solution 2#
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| class Solution:
def countNumbersWithUniqueDigits(self, n: int) -> int:
@cache
def dfs(pos, mask, lead):
if pos <= 0:
return 1
ans = 0
for i in range(10):
if (mask >> i) & 1:
continue
if i == 0 and lead:
ans += dfs(pos - 1, mask, lead)
else:
ans += dfs(pos - 1, mask | (1 << i), False)
return ans
return dfs(n, 0, True)
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| class Solution {
private int[][] dp = new int[10][1 << 11];
public int countNumbersWithUniqueDigits(int n) {
for (var e : dp) {
Arrays.fill(e, -1);
}
return dfs(n, 0, true);
}
private int dfs(int pos, int mask, boolean lead) {
if (pos <= 0) {
return 1;
}
if (!lead && dp[pos][mask] != -1) {
return dp[pos][mask];
}
int ans = 0;
for (int i = 0; i < 10; ++i) {
if (((mask >> i) & 1) == 1) {
continue;
}
if (i == 0 && lead) {
ans += dfs(pos - 1, mask, lead);
} else {
ans += dfs(pos - 1, mask | (1 << i), false);
}
}
if (!lead) {
dp[pos][mask] = ans;
}
return ans;
}
}
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| class Solution {
public:
int dp[10][1 << 11];
int countNumbersWithUniqueDigits(int n) {
memset(dp, -1, sizeof dp);
return dfs(n, 0, true);
}
int dfs(int pos, int mask, bool lead) {
if (pos <= 0) {
return 1;
}
if (!lead && dp[pos][mask] != -1) {
return dp[pos][mask];
}
int ans = 0;
for (int i = 0; i < 10; ++i) {
if ((mask >> i) & 1) continue;
if (i == 0 && lead) {
ans += dfs(pos - 1, mask, lead);
} else {
ans += dfs(pos - 1, mask | 1 << i, false);
}
}
if (!lead) {
dp[pos][mask] = ans;
}
return ans;
}
};
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| func countNumbersWithUniqueDigits(n int) int {
dp := make([][]int, 10)
for i := range dp {
dp[i] = make([]int, 1<<11)
for j := range dp[i] {
dp[i][j] = -1
}
}
var dfs func(int, int, bool) int
dfs = func(pos, mask int, lead bool) int {
if pos <= 0 {
return 1
}
if !lead && dp[pos][mask] != -1 {
return dp[pos][mask]
}
ans := 0
for i := 0; i < 10; i++ {
if ((mask >> i) & 1) == 1 {
continue
}
if i == 0 && lead {
ans += dfs(pos-1, mask, lead)
} else {
ans += dfs(pos-1, mask|1<<i, false)
}
}
if !lead {
dp[pos][mask] = ans
}
return ans
}
return dfs(n, 0, true)
}
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