# Rotated Digits Problem

## Description

LeetCode Problem 788.

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:

``````1
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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:

``````1
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Input: n = 1
Output: 0
``````

Example 3:

``````1
2
Input: n = 2
Output: 1
``````

Constraints:

• 1 <= n <= 10^4

## Sample C++ Code

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class Solution {
public:
int rotatedDigits(int N) {
int count = 0;
int num, num_r;
int rot = {0, 1, 5, -1, -1, 2, 9, -1, 8, 6};

for (int i = 0; i < 10; i ++) {
for (int j = 0; j < 10; j ++) {
for (int k = 0; k < 10; k ++) {
for (int l = 0; l < 10; l ++) {
num = 1000 * i + 100 * j + 10 * k + l;
if (num > N) {
return count;
}
if ((rot[i] == -1) || (rot[j] == -1) || (rot[k] == -1) || (rot[l] == -1)) {
continue;
}
num_r = 1000 * rot[i] + 100 * rot[j] + 10 * rot[k] + rot[l];
if (num != num_r)
count ++;
}
}
}
}
return count;
}
};
``````