Can someone please help me write a shell script for this??

Write a shell script file_sizes.sh which prints the names of the files in the current directory splitting them into three categories: small if the file contains less than 10 lines medium-sized if the file contains between 10 and 100 lines large if the file contains 100 or more lines . Your script should always print exactly three lines of output. Files should be listed in alphabetic order on each line. Your shell-script should match character-for-character the output shown in the example below. Notice the creation of a separate directory for testing and the use of the script from the last question to produce test files. You could also produce test files manually using an editor. $ mkdir test $ cd test $ ../echon.sh 5 text > a $ ../echon.sh 505 text > bbb $ ../echon.sh 17 text > cc $ ../echon.sh 10 text > d $ ../echon.sh 1000 text > e $ ../echon.sh o text > empty $ ls -1 total 24 -rw-r--r-- 1 andrewt andrewt 25 Mar 24 10:37 a -rw-r--r-- 1 andrewt andrewt 2525 Mar 24 10:37 bbb -rw-r--r-- 1 andrewt andrewt 85 Mar 24 10:37 CC -rw-r--r-- 1 andrewt andrewt 50 Mar 24 10:37 d -rw-r--r-- 1 andrewt andrewt 5000 Mar 24 10:37 e -rw-r--r-- 1 andrewt andrewt Mar 24 10:37 empty $ ../file_sizes.sh Small files: a empty Medium-sized files: cod Large files: bbb e HINT: You can use the command wc to discover how many lines are in a file. You probably want to use a subshell $(), if statements, and the [ (test) command