# LeetCode 1056. Confusing Number

## Description

https://leetcode.com/problems/confusing-number/

Given a number `N`, return `true` if and only if it is a confusing number, which satisfies the following condition:

We can rotate digits by 180 degrees to form new digits. When 0, 1, 6, 8, 9 are rotated 180 degrees, they become 0, 1, 9, 8, 6 respectively. When 2, 3, 4, 5 and 7 are rotated 180 degrees, they become invalid. A confusing number is a number that when rotated 180 degrees becomes a different number with each digit valid.

Example 1:

```Input: 6
Output: true
Explanation:
We get `9` after rotating `6`, `9` is a valid number and `9!=6`.
```

Example 2:

```Input: 89
Output: true
Explanation:
We get `68` after rotating `89`, `86` is a valid number and `86!=89`.
```

Example 3:

```Input: 11
Output: false
Explanation:
We get `11` after rotating `11`, `11` is a valid number but the value remains the same, thus `11` is not a confusing number.
```

Example 4:

```Input: 25
Output: false
Explanation:
We get an invalid number after rotating `25`.
```

Note:

1. `0 <= N <= 10^9`
2. After the rotation we can ignore leading zeros, for example if after rotation we have `0008` then this number is considered as just `8`.

## Explanation

Create a mapping between numbers and their rotated numbers and check if the reversed string can be same as their rotated string.

## Python Solution

``````class Solution:
def confusingNumber(self, N: int) -> bool:
mapping = {
'0': '0',
'1': '1',
'6': '9',
'8': '8',
'9': '6',
}

rotated_str = ""
n_str = str(N)
for c in n_str[::-1]:
if c not in mapping:
return False
else:
rotated_str += mapping[c]

return rotated_str != n_str
``````
• Time Complexity: O(N).
• Space Complexity: O(N).