Checking if digits in a string are symmetric (in edges)

Question

I have a string of numbers. I need to check if the numbers on the edges are symmetric, meaning they have the same remainder when modulo by 2.

I have written a code which works, but I have something troubling my mind about that, after some failures I've come up with this code:

int PaliPair(char* st, int n)
{
  if(n<=1) return 1;
  return (*st%2 == *(st+n-1)%2) && PaliPair(st +1, n-2);
}

The question is, why do I have to return n-2 and not n-1? I'm kinda confused of why it works. Any explanation would be highly appreciated. I think I'm missing something, perhaps the fact that the string ends with "\0" which I need to conclude from that something.

Vlad from Moscow · Accepted Answer · 2016-03-01T11:26:57.970

If you have a string for example like this

"1243"

then you at first check the first and the last characters.

Then you need to check the characters in the middle that is

"24"

So the target string now has length 4 - 2 (the number of characters that were already checked)

So in each recursion you check 2 characters, In the next recursion you need to check 2 less characters.

As for the function itself I would write it like

int PaliPair( const char *s, size_t n )
{
   return n < 2 || *s % 2 == *( s + n - 1 ) % 2 && PaliPair( s + 1, n - 2 );
}

Or even like

int PaliPair( const char *s, size_t n )
{
   return n < 2 || ( *s - '0' ) % 2 == ( *( s + n - 1 ) - '0' ) % 2 && PaliPair( s + 1, n - 2 );
}

score 2 · Answer 2 · answered Mar 01 '16 at 10:57

Suppose your string is 21312, at first step you will compare 2 and 2 from both ends. Then you go one step forward with st+1 you should consider 131, so you should not consider 2's from beginning and the end of your string, that's why you do n-2.

In this way you go one character forward from beginning of the string but you should also shift one character backward from the end of the string as well. I hope you get my point.

Checking if digits in a string are symmetric (in edges)

2 Answers2