How to find the length of the longest substring not containing 0

[ak2]

Hi! I have a string something like this: "1234000111123012001210", i.e. it consists of random natural numbers between 0 and 9 and max lenght is 24. I need to find out how long the longest substring not containing 0 is. In this example the answer is 6 because the substring "111123" is the longest one not containing a 0. How can I do this? I was thinking of putting my string in an array and then loop thru it and count the characters until a 0 occurs, keep the value and compare it against the next substring not containing 0. My thoughts in code:

data a;

str='1234000111123012001210';

array b[24] $;

do i=1 to dim(b);

b[i]=substr(str,i,1);

end;

drop i;

count=0;/*the counter used to find the longest substring*/

do i=1 to dim(b);

if b[i]>0 then count=count+1; /*this should caputre 1234 and count=4*/

else /*keep the value 4 and do not count the preceding three 0*/

/*then go back to counting the substring '111123'.count=6. compare this count with count=4. keep count=6 */

end;

drop i;

run;

I am stuck and would appreciate it if anyone can provide me with some help.

[LinusH]

You need two count variables, one that you increment continuously, and one that holds the current maximum count.

Another totally different two-step approach would be to first generate a new observation for each substring. Then you could do a subsequent SQL using max(length()) construct.

[Ksharp · Accepted solution]

data _null_;
str='1234000111123012001210';
do i=1 to countw(str,'0');
 max=max(max,length(scan(str,i,'0')));
end;
putlog "max:" max;
run;

[ak2]

This was exactly what I needed! Thanks a lot!

[RW9]

Simply split the string by "0", and count length each time retaining a maximum:

data want (drop=i);
  str="1234000111123012001210";
  i=1;
  max_len=0;
  do while(scan(str,i,"0") ne "");
    max_len=max(lengthn(scan(str,i,"0")),max_len); 
    i=i+1;
  end;
run;

[ak2]

This is what I needed. Thanks a lot! Case closed.

[RW9]

Please mark @Ksharp 's post as the answer.