## Algorithm

Problem Name: 788. Rotated Digits

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`, and `8` rotate to themselves,
• `2` and `5` rotate to each other (in this case they are rotated in a different direction, in other words, `2` or `5` gets mirrored),
• `6` and `9` rotate 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`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int rotatedDigits(int n) {
int[] dp = new int[n + 1];
int result = 0;
for (int i = 0; i <= Math.min(n, 9); i++) {
if (i == 0 || i == 1 || i == 8) {
dp[i] = 1;
} else if (i == 2 || i == 5 || i == 6 || i == 9) {
dp[i] = 2;
result++;
}
}
for (int i = 10; i <= n; i++) {
int factor = dp[i / 10];
int remainder = dp[i % 10];
if (factor == 1 && remainder == 1) {
dp[i] = 1;
} else if (factor >= 1 && remainder >= 1) {
dp[i] = 2;
result++;
}
}
return result;
}
}
``````
Copy The Code &

Input

cmd
n = 10

Output

cmd
4

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const rotatedDigits = function(n) {
const dp = new Array(n + 1).fill(0);
let count = 0;
for(let i = 0; i <= n; i++){
if(i < 10){
if(i == 0 || i == 1 || i == 8) dp[i] = 1;
else if(i == 2 || i == 5 || i == 6 || i == 9){
dp[i] = 2;
count++;
}
} else {
let a = dp[~~(i / 10)], b = dp[i % 10];
if(a == 1 && b == 1) dp[i] = 1;
else if(a >= 1 && b >= 1){
dp[i] = 2;
count++;
}
}
}
return count;
};
``````
Copy The Code &

Input

cmd
n = 10

Output

cmd
4

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def rotatedDigits(self, N):
"""
:type N: int
:rtype: int
"""
res = 0
for i in range(1, N + 1):
i = str(i)
tmp = []
check = True
for char in i:
if char in ("3", "4", "7"):
check = False
break
if char in ("0", "1", "8"):
tmp.append(char)
if char == "2":
tmp.append("5")
if char == "5":
tmp.append("2")
if char == "6":
tmp.append("9")
if char == "9":
tmp.append("6")
if check and i != "".join(tmp): res += 1
return res
``````
Copy The Code &

Input

cmd
n = 1

### #4 Code Example with C# Programming

```Code - C# Programming```

``````
using System.Collections.Generic;

namespace LeetCode
{
public class _0788_RotatedDigits
{
public int RotatedDigits(int N)
{
var num = N.ToString();
var length = num.Length;

var memo = new Dictionary<(int length, bool equality_flag, bool involution_flag), int>();
return DP(0, true, false, num, length, memo);
}

public int DP(int k, bool equality_flag, bool involution_flag, string num, int length, IDictionary<(int length, bool equality_flag, bool involution_flag), int> memo)
{
if (k == length) return involution_flag ? 1 : 0;
if (!memo.ContainsKey((k, equality_flag, involution_flag)))
{
var current = 0;
for (char ch = '0'; ch <= (equality_flag ? num[k] : '9'); ch++)
{
if (ch == '3' || ch == '4' || ch == '7') continue;
current += DP(k + 1, equality_flag && ch == num[k], involution_flag || (ch == '2' || ch == '5' || ch == '6' || ch == '9'), num, length, memo);
}

memo[(k, equality_flag, involution_flag)] = current;
}

return memo[(k, equality_flag, involution_flag)];
}
}
}
``````
Copy The Code &

Input

cmd
n = 1