An integer x is a good if after rotating each digit individually by 180 degrees, we get a valid number that is different from x. Each digit must be rotated - we cannot choose to leave it alone.
A number is valid if each digit remains a digit after rotation. For example:
0,1, and8rotate to themselves,2and5rotate to each other (in this case they are rotated in a different direction, in other words,2or5gets mirrored),6and9rotate to each other, and- the rest of the numbers do not rotate to any other number and become invalid.
Given an integer n, return the number of good integers in the range [1, n].
Example 1:
Input: n = 10 Output: 4 Explanation: There are four good numbers in the range [1, 10] : 2, 5, 6, 9. Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.
Example 2:
Input: n = 1 Output: 0
Example 3:
Input: n = 2 Output: 1
Constraints:
1 <= n <= 104
class Solution:
def rotatedDigits(self, n: int) -> int:
def check(x):
y, t = 0, x
k = 1
while t:
v = t % 10
if d[v] == -1:
return False
y = d[v] * k + y
k *= 10
t //= 10
return x != y
d = [0, 1, 5, -1, -1, 2, 9, -1, 8, 6]
return sum(check(i) for i in range(1, n + 1))class Solution:
def rotatedDigits(self, n: int) -> int:
@cache
def dfs(pos, ok, limit):
if pos <= 0:
return ok
up = a[pos] if limit else 9
ans = 0
for i in range(up + 1):
if i in (0, 1, 8):
ans += dfs(pos - 1, ok, limit and i == up)
if i in (2, 5, 6, 9):
ans += dfs(pos - 1, 1, limit and i == up)
return ans
a = [0] * 6
l = 1
while n:
a[l] = n % 10
n //= 10
l += 1
return dfs(l, 0, True)class Solution {
private int[] d = new int[] {0, 1, 5, -1, -1, 2, 9, -1, 8, 6};
public int rotatedDigits(int n) {
int ans = 0;
for (int i = 1; i <= n; ++i) {
if (check(i)) {
++ans;
}
}
return ans;
}
private boolean check(int x) {
int y = 0, t = x;
int k = 1;
while (t > 0) {
int v = t % 10;
if (d[v] == -1) {
return false;
}
y = d[v] * k + y;
k *= 10;
t /= 10;
}
return x != y;
}
}class Solution {
private int[] a = new int[6];
private int[][] dp = new int[6][2];
public int rotatedDigits(int n) {
int len = 0;
for (var e : dp) {
Arrays.fill(e, -1);
}
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true);
}
private int dfs(int pos, int ok, boolean limit) {
if (pos <= 0) {
return ok;
}
if (!limit && dp[pos][ok] != -1) {
return dp[pos][ok];
}
int up = limit ? a[pos] : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (i == 0 || i == 1 || i == 8) {
ans += dfs(pos - 1, ok, limit && i == up);
}
if (i == 2 || i == 5 || i == 6 || i == 9) {
ans += dfs(pos - 1, 1, limit && i == up);
}
}
if (!limit) {
dp[pos][ok] = ans;
}
return ans;
}
}class Solution {
public:
const vector<int> d = {0, 1, 5, -1, -1, 2, 9, -1, 8, 6};
int rotatedDigits(int n) {
int ans = 0;
for (int i = 1; i <= n; ++i) {
ans += check(i);
}
return ans;
}
bool check(int x) {
int y = 0, t = x;
int k = 1;
while (t) {
int v = t % 10;
if (d[v] == -1) {
return false;
}
y = d[v] * k + y;
k *= 10;
t /= 10;
}
return x != y;
}
};class Solution {
public:
int a[6];
int dp[6][2];
int rotatedDigits(int n) {
memset(dp, -1, sizeof dp);
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true);
}
int dfs(int pos, int ok, bool limit) {
if (pos <= 0) {
return ok;
}
if (!limit && dp[pos][ok] != -1) {
return dp[pos][ok];
}
int up = limit ? a[pos] : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (i == 0 || i == 1 || i == 8) {
ans += dfs(pos - 1, ok, limit && i == up);
}
if (i == 2 || i == 5 || i == 6 || i == 9) {
ans += dfs(pos - 1, 1, limit && i == up);
}
}
if (!limit) {
dp[pos][ok] = ans;
}
return ans;
}
};func rotatedDigits(n int) int {
d := []int{0, 1, 5, -1, -1, 2, 9, -1, 8, 6}
check := func(x int) bool {
y, t := 0, x
k := 1
for ; t > 0; t /= 10 {
v := t % 10
if d[v] == -1 {
return false
}
y = d[v]*k + y
k *= 10
}
return x != y
}
ans := 0
for i := 1; i <= n; i++ {
if check(i) {
ans++
}
}
return ans
}func rotatedDigits(n int) int {
a := make([]int, 6)
dp := make([][2]int, 6)
for i := range a {
dp[i] = [2]int{-1, -1}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}
var dfs func(int, int, bool) int
dfs = func(pos, ok int, limit bool) int {
if pos <= 0 {
return ok
}
if !limit && dp[pos][ok] != -1 {
return dp[pos][ok]
}
up := 9
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
if i == 0 || i == 1 || i == 8 {
ans += dfs(pos-1, ok, limit && i == up)
}
if i == 2 || i == 5 || i == 6 || i == 9 {
ans += dfs(pos-1, 1, limit && i == up)
}
}
if !limit {
dp[pos][ok] = ans
}
return ans
}
return dfs(l, 0, true)
}