Extracting the user with the most amount of files in a dir

I am currently working on a script that should receive a standard input, and output the user with the highest amount of files in that directory.

I've wrote this so far:

#!/bin/bash 
while read DIRNAME
do
        ls -l $DIRNAME | awk 'NR>1 {print $4}' | uniq -c      
done

and this is the output I get when I enter /etc for an instance:

 26 root
  1 dip
  8 root
  1 lp
 35 root
  2 shadow
 81 root
  1 dip
 27 root
  2 shadow
 42 root

Now obviously the root folder is winning in this case, but I don't want only to output this, i also want to sum the number of files and output only the user with the highest amount of files.

Expected output for entering /etc:

root

is there a simple way to filter the output I get now, so that the user with the highest sum will be stored somehow?

ls -l /etc | awk 'BEGIN{FS=OFS=" "}{a[$4]+=1}END{ for (i in a) print a[i],i}' | sort -g -r | head -n 1 | cut -d' ' -f2

This snippet returns the group with the highest number of files in the /etc directory.

What it does:

1. ls -l /etc lists all the files in /etc in long form.
2. awk 'BEGIN{FS=OFS=" "}{a[$4]+=1}END{ for (i in a) print a[i],i}' sums the number of occurrences of unique words in the 4th column and prints the number followed by the word.
3. sort -g -r sorts the output descending based on numbers.
4. head -n 1 takes the first line
5. cut -d' ' -f2 takes the second column while the delimiter is a white space.

Note: In your question, you are saying that you want the user with the highest number of files, but in your code you are referring to the 4th column which is the group. My code follows your code and groups on the 4th column. If you wish to group by user and not group, change {a[$4]+=1} to {a[$3]+=1}.