How can I count most occuring sequence of 3 letters within a word with a bash script

Question

I have a sample file like

XYZAcc
ABCAccounting
Accounting firm
Accounting Aco
Accounting Acompany
Acoustical consultant

Here I need to grep

Answer 1

This is an alternative method to the solution of Ed Morton. It is less looping, but needs a bit more memory. The idea is not to care about spaces or any non-alphabetic character. We filter them out in the end.

awk -v n=3 '{ for(i=length-n+1;i>0;--i) a[tolower(substr($0,i,n))]++ }
            END {for(s in a) if (s !~ /[^a-z]/) print s,a[s] }' file

When you use GNU awk, you can do this a bit differently and optimized by setting each record to be a word. This way the end selection does not need to happen:

awk -v n=3 -v RS='[[:space:]]' '
    (length>=n){ for(i=length-n+1;i>0;--i) a[tolower(substr($0,i,n))]++ }
    END {for(s in a) print s,a[s] }' file