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From: Ben Bacarisse <ben.usenet@bsb.me.uk>
Newsgroups: comp.programming
Subject: Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post)
Date: Thu, 27 Feb 2020 15:55:25 +0000
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"Chad Altenburg, Ph.D." <cdalten@gmail.com> writes:

> And here is the problem statement... 
>
> Look for patterns like "zip" and "zap" in the string -- length-3,
> starting with 'z' and ending with 'p'. Return a string where for all
> such words, the middle letter is gone, so "zipXzap" yields "zpXzp".
>
> zipZap("zipXzap") → "zpXzp" 
> zipZap("zopzop") → "zpzp" 
> zipZap("zzzopzop") → "zzzpzp" 
>
>
> But the solution that I somehow managed to come up with when I was
> drunk, stoned, depressed, and pissed off, is well, ugly...
>
> Here is the solution 
>
> public String zipZap(String str) {
>   String newString = "";
>   
>   if (str.length() < 3) {
>     return str;
>   }
>   
>   for (int i = 0; i < str.length() - 3; i++) {
>     if (str.charAt(i) == 'z' && str.charAt(i + 2) == 'p' ) {
>       newString = newString +  "zp";
>       i = i + 2;
>     }
>     else {
>       newString = newString + str.charAt(i);
>     }
>     
>   }
>   
>   for (int i = str.length() - 3; i < str.length(); i++ ) {
>     if (str.charAt(i) == 'z' && str.charAt(i + 2) == 'p' ) {
>       newString = newString +  "zp";
>       i = i + 2;
>     }
>     else {
>       newString = newString + str.charAt(i);
>     }
>   }
>   
>   return newString;
> }
>
> The problem is that I have two loops. This is because I didn't know
> how to attack the case when I have to search for a pattern at the
> *end* of the string. If I use str.length(), I go out of bounds. So I
> ended up using a second loop to handle the end of the string case.
>
> My gut feeling says this isn't right. But I have no idea how this
> could be done in one loop.

Here's a C solution.  It's messy because the natural way to do string
things in C is to pass an array to take the result.  You then need to
pass a size so the function does not write out-of-bounds.  But to be
useful the function should keep going and return the size that's needed
even when there was no room this time:

#include <stdio.h>

size_t zipzap(const char *src, char *dst, size_t n)
{
     size_t i = 0;
     while (*src) {
          if (i < n)
               dst[i] = *src;
          i += 1;
          if (src[0] == 'z' && src[1] && src[2] == 'p') {
               if (i < n)
                    dst[i] = 'p';
               i += 1;
               src += 3;
          }
          else src += 1;
     }
     dst[i <= n ? i : n-1] = 0;
     return i + 1;
}

int main(int argc, char **argv)
{
     char z[10];
     for (int a = 1; a < argc; a++) {
          size_t required = zipzap(argv[a], z, sizeof z);
          printf("%s -> %s (%zu)\n", argv[a], z, required);
     }
}

The more "scripty" way -- allocating the result -- looks simpler:

char *zipzap_str(const char *src)
{
     char result[strlen(src) + 1], *dst = result;
     while (*src) {
          *dst++ = *src++;
          if (src[-1] == 'z' && src[0] && src[1] == 'p') {
               *dst++ = 'p';
               src += 2;
          }
     }
     *dst++ = 0;
     char *rp = malloc(dst - result);
     return rp ? strcpy(rp, result) : 0;
}

-- 
Ben.