Given an integer n, return the number of positive integers in the range [1, n] that have at least one repeated digit.
Example 1:
Input: n = 20 Output: 1 Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.
Example 2:
Input: n = 100 Output: 10 Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
Example 3:
Input: n = 1000 Output: 262
Constraints:
1 <= n <= 109
class Solution:
def numDupDigitsAtMostN(self, n: int) -> int:
return n - self.f(n)
def f(self, n):
def A(m, n):
return 1 if n == 0 else A(m, n - 1) * (m - n + 1)
vis = [False] * 10
ans = 0
digits = [int(c) for c in str(n)[::-1]]
m = len(digits)
for i in range(1, m):
ans += 9 * A(9, i - 1)
for i in range(m - 1, -1, -1):
v = digits[i]
j = 1 if i == m - 1 else 0
while j < v:
if not vis[j]:
ans += A(10 - (m - i), i)
j += 1
if vis[v]:
break
vis[v] = True
if i == 0:
ans += 1
return ansclass Solution:
def numDupDigitsAtMostN(self, n: int) -> int:
return n - self.f(n)
def f(self, n):
@cache
def dfs(pos, mask, lead, limit):
if pos <= 0:
return lead ^ 1
up = a[pos] if limit else 9
ans = 0
for i in range(up + 1):
if (mask >> i) & 1:
continue
if i == 0 and lead:
ans += dfs(pos - 1, mask, lead, limit and i == up)
else:
ans += dfs(pos - 1, mask | 1 << i, False, limit and i == up)
return ans
a = [0] * 11
l = 0
while n:
l += 1
a[l] = n % 10
n //= 10
return dfs(l, 0, True, True)class Solution {
public int numDupDigitsAtMostN(int n) {
return n - f(n);
}
public int f(int n) {
List<Integer> digits = new ArrayList<>();
while (n != 0) {
digits.add(n % 10);
n /= 10;
}
int m = digits.size();
int ans = 0;
for (int i = 1; i < m; ++i) {
ans += 9 * A(9, i - 1);
}
boolean[] vis = new boolean[10];
for (int i = m - 1; i >= 0; --i) {
int v = digits.get(i);
for (int j = i == m - 1 ? 1 : 0; j < v; ++j) {
if (vis[j]) {
continue;
}
ans += A(10 - (m - i), i);
}
if (vis[v]) {
break;
}
vis[v] = true;
if (i == 0) {
++ans;
}
}
return ans;
}
private int A(int m, int n) {
return n == 0 ? 1 : A(m, n - 1) * (m - n + 1);
}
}class Solution {
private int[] a = new int[11];
private int[][] dp = new int[11][1 << 11];
public int numDupDigitsAtMostN(int n) {
return n - f(n);
}
private int f(int n) {
for (var e : dp) {
Arrays.fill(e, -1);
}
int len = 0;
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true, true);
}
private int dfs(int pos, int mask, boolean lead, boolean limit) {
if (pos <= 0) {
return lead ? 0 : 1;
}
if (!lead && !limit && dp[pos][mask] != -1) {
return dp[pos][mask];
}
int up = limit ? a[pos] : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (((mask >> i) & 1) == 1) {
continue;
}
if (i == 0 && lead) {
ans += dfs(pos - 1, mask, lead, limit && i == up);
} else {
ans += dfs(pos - 1, mask | 1 << i, false, limit && i == up);
}
}
if (!lead && !limit) {
dp[pos][mask] = ans;
}
return ans;
}
}class Solution {
public:
int numDupDigitsAtMostN(int n) {
return n - f(n);
}
int f(int n) {
int ans = 0;
vector<int> digits;
while (n) {
digits.push_back(n % 10);
n /= 10;
}
int m = digits.size();
vector<bool> vis(10);
for (int i = 1; i < m; ++i) {
ans += 9 * A(9, i - 1);
}
for (int i = m - 1; ~i; --i) {
int v = digits[i];
for (int j = i == m - 1 ? 1 : 0; j < v; ++j) {
if (!vis[j]) {
ans += A(10 - (m - i), i);
}
}
if (vis[v]) {
break;
}
vis[v] = true;
if (i == 0) {
++ans;
}
}
return ans;
}
int A(int m, int n) {
return n == 0 ? 1 : A(m, n - 1) * (m - n + 1);
}
};class Solution {
public:
int a[11];
int dp[11][1 << 11];
int numDupDigitsAtMostN(int n) {
return n - f(n);
}
int f(int n) {
memset(dp, -1, sizeof dp);
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true, true);
}
int dfs(int pos, int mask, bool lead, bool limit) {
if (pos <= 0) {
return lead ? 0 : 1;
}
if (!lead && !limit && dp[pos][mask] != -1) {
return dp[pos][mask];
}
int up = limit ? a[pos] : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if ((mask >> i) & 1) continue;
if (i == 0 && lead) {
ans += dfs(pos - 1, mask, lead, limit && i == up);
} else {
ans += dfs(pos - 1, mask | 1 << i, false, limit && i == up);
}
}
if (!lead && !limit) {
dp[pos][mask] = ans;
}
return ans;
}
};func numDupDigitsAtMostN(n int) int {
return n - f(n)
}
func f(n int) int {
digits := []int{}
for n != 0 {
digits = append(digits, n%10)
n /= 10
}
m := len(digits)
vis := make([]bool, 10)
ans := 0
for i := 1; i < m; i++ {
ans += 9 * A(9, i-1)
}
for i := m - 1; i >= 0; i-- {
v := digits[i]
j := 0
if i == m-1 {
j = 1
}
for ; j < v; j++ {
if !vis[j] {
ans += A(10-(m-i), i)
}
}
if vis[v] {
break
}
vis[v] = true
if i == 0 {
ans++
}
}
return ans
}
func A(m, n int) int {
if n == 0 {
return 1
}
return A(m, n-1) * (m - n + 1)
}func numDupDigitsAtMostN(n int) int {
return n - f(n)
}
func f(n int) int {
a := make([]int, 11)
dp := make([][]int, 11)
for i := range dp {
dp[i] = make([]int, 1<<11)
for j := range dp[i] {
dp[i][j] = -1
}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}
var dfs func(int, int, bool, bool) int
dfs = func(pos, mask int, lead, limit bool) int {
if pos <= 0 {
if lead {
return 0
}
return 1
}
if !lead && !limit && dp[pos][mask] != -1 {
return dp[pos][mask]
}
ans := 0
up := 9
if limit {
up = a[pos]
}
for i := 0; i <= up; i++ {
if ((mask >> i) & 1) == 1 {
continue
}
if i == 0 && lead {
ans += dfs(pos-1, mask, lead, limit && i == up)
} else {
ans += dfs(pos-1, mask|1<<i, false, limit && i == up)
}
}
if !lead && !limit {
dp[pos][mask] = ans
}
return ans
}
return dfs(l, 0, true, true)
}