## Interview Question

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Country: United States

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There is no middle of anything when we are looking at a number. A number, rather more precisely integer is given by this regex (ignoring sign) :

``N -> [0-9]+``

That would mean, in a context free way, it can be written as :

``N --> [0-9]* [0-9]? [0-9]*``

That is how you should define middle. In this definition, of course middle becomes meaningless. Example: 34543 . What is the middle? My grammar of course is terribly ambiguous, but hope I did make the point.

Now, of course there are two problems.
1. Get all the factors of a number. This is not very interesting problem.
[ geeksforgeeks.org/find-divisors-natural-number-set-1/ ]
2. A predicate to declare if a number is a special number. That is slightly more interesting.
But then fairly trivially done by (with proper collection minus and subset definition ):

``````special_digits = '356'.toCharArray
all_digits = ['0':'9'].string.toCharArray
drop_digits = all_digits - special_digits

def is_special_no( n ){
// set minus : to get the left over digits after subtracting drop_digits
left =  set ( str(n).toCharArray ) - drop_digits
// when left is non empty and is a proper subset of special_digits then true
!empty(left) && left <= special_digits
}

println( is_special_no(104) )``````

But the bigger question is, why write this way? It is terribly un-optimal !Clearly there are better way to do so? As it turned out, there is!!

``````special_digits = set ( '356'.toCharArray )
def is_special_no( n ){
!empty ( select ( str( n ).toCharArray ) where { \$.o @ special_digits } )
}

println( is_special_no(134) )``````

Now this is cleaner, neater, and optimal. Now it is trivial to generate the appropriate imperative code.

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#include<stdio.h>
int main()
{
long long num,count=0;int i;
scanf("%lld",&num);
for(i=2;i*i<=num;i++)
{
if(num%i==0)
{
int factor2=num/i;
int k=i;
while(k)
{
int t=k%10;k/=10;
if(t==3||t==5||t==6){count++;break;}
}
if(k!=factor2)
{
while(factor2)
{
int t=factor2%10;factor2/=10;
if(t==3||t==5||t==6){count++;break;}
}
}
}
}
while(num)
{
int t=num%10;
num/=10;
if(t==3||t==5||t==6){count++;break;}
}
printf("%lld\n",count);

}

Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
int main()
{
long long num,count=0;int i;
scanf("%lld",&num);
for(i=2;i*i<=num;i++)
{
if(num%i==0)
{
int factor2=num/i;
int k=i;
while(k)
{
int t=k%10;k/=10;
if(t==3||t==5||t==6){count++;break;}
}
if(k!=factor2)
{
while(factor2)
{
int t=factor2%10;factor2/=10;
if(t==3||t==5||t==6){count++;break;}
}
}
}
}
while(num)
{
int t=num%10;
num/=10;
if(t==3||t==5||t==6){count++;break;}
}
printf("%lld\n",count);

}

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