(Medium) Good Number LeetCode - Mock Interview

Description

X is a good number 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. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.

Now given a positive number N, how many numbers X from 1 to N are good?

Example:
Input: 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.

Note:

• N  will be in range [1, 10000].

Solution

class Solution {
    public int rotatedDigits(int N) {
        
        int count = 0; 
        
        for(int i =2;i<=N;i++){
            
            if(Judge(i)){
                count++;
            }
            
        }
        return count;
        
    }
    
    
    public Boolean Judge(int N){
        Boolean b = true;
        int res=N; 
        int dig=0; 
        int NewNum =0;
        int  t = 1;
        
        if (N<=10){
           do{
                dig = res%10;
                res = res/10;



                if(dig ==1 ||dig ==0 ||dig ==3 ||dig==4 ||dig ==7 ||dig ==8 )
                    return false;
            }

            while(
                res>0
            );

                return true;
            
        }
        
        else{
               do {
                    dig = res%10;
                    res = res/10;

                if(dig ==3 ||dig==4 ||dig ==7 ) {
                      return false;
                    
                }
                  
                else if( dig ==2) {
                    dig = 5;
                }
                else if(dig ==5) {
                    dig =2;
                }
                else if (dig  ==6) {
                    dig = 9;
                }
                else if(dig ==9) {
                    dig =6;
                }
                    
                    NewNum = NewNum + dig*t;
                    
                    t = t*10;
                }
                
              while(
                res>0
            );
            
            if(NewNum !=N){
                return true;
            }
            else{
                return false;
            }
            
        }
       
      
    }
}