UNIX for Dummies Questions & Answers

$ ls
ovwbemaction_lib.vcproj  OvWBEMAction_lib.vcproj

ls | awk '!a[toupper($0)] {a[toupper($0)]=$0;next}{print a[toupper($0)] RS $0}'

I have directories like, 

$ ls -1 -R I18N_tutorial/
I18N_tutorial/:
Converting MS VC++ project to use Msgs.html
Converting to Use Msgs - C-C++ Example.htm
Converting to Use Msgs - C-C++ Example.url
nls
pre-nls
readme.txt
src

I18N_tutorial/nls:
C
de_DE.roman8

I18N_tutorial/nls/C:
Imakefile

I18N_tutorial/nls/de_DE.roman8:
hello.1
Imakefile

I18N_tutorial/pre-nls:
hello.1
hello.sets

I18N_tutorial/src:
hello.c
hello.C
hello.dsp
helloMsg.c
Imakefile
name.c
name.C

Out of this, my expectation is to find out only name.c and hello.c because it has the same name but different case. 

so I executed the below command,
$ ls -1 -R  I18N_tutorial/ | awk '!a[toupper($0)] {a[toupper($0)]=$0;next}{print a[toupper($0)] RS $0}'
Imakefile
Imakefile
hello.1
hello.1
hello.c
hello.C
Imakefile
Imakefile
name.c
name.C

Its giving the above output. 

but If I issue the command, with in the src directory it is working fine. 

$ cd I18N_tutorial/src/
$ ls
hello.c  hello.C  hello.dsp  helloMsg.c  Imakefile  name.c  name.C
$ ls  -1 . | awk '!a[toupper($0)] {a[toupper($0)]=$0;next}{print a[toupper($0)] RS $0}'
hello.c
hello.C
name.c
name.C

Its giving the output what i want exactly.

find . | tr A-Z a-z | sort | uniq -d | while read f; do find "`dirname "$f"`" -maxdepth 1 -iname "`basename "$f"`"; done