bash - Find max number of concurrent events -

i'd print max number of concurrent events given start time , end time of each event in "hhmm" format (example input below)

$ cat input.txt   1030,1100  1032,1100 1032,1033  1033,1050  1034,1054 1039,1043  1040,1300 

for this, would

• sort start time (column 1)
• use awk/sed iterate on values in column 2 (i.e end time) find count of end times preceeding event greater current value (i.e find running events). elaborate, assuming line 3 being processed awk ... end time 10:33. end times of preceding 2 events 11:00 , 11:00. since both these values greater 10:33 (i.e. still running @ 10:33), third column (i.e. number of concurrent jobs) contain 2 line

the expected output of awk script find concurrent events input be

0 1 2 2 2 4 0  

• find max value of third column.

my awk rudimentary @ best , having difficulty implementing step 2. i'd pure script without resorting heavy weight language java. hence awk gurus highly appreciated. non-awk linux 1 liners welcome.

begin  {fs="\,"; i=0}  { superpos=0;     (j=1; j<=i; j++ ){       if($2 < a[j,2])           ++superpos      }       a[++i,1]=$1;       a[i,2]=$2;       print superpos;       a[i,3]=superpos; } end{ max=0;          (j=1; j<=i; j++ ){            if ( a[j,3]>max)             max= a[j,3];      }      print "max = ",max;  } 

running @ ideone

hth!

output:

0 0 2 2 2 4 0 max =  4 

edit

or more awkish, if prefer:

begin  {fs="\,"; max=0 } {      b=0;      (var in a){       if($2 < a[var]) b++;     }         a[nr]=$2;       print b;       if (b > max) max = b; } end { print "max = ", max }