Internal table record checking (Urgent)

Hi Experts,
I have an internal table having records. I want to check each record of the table with every other record to find out if there exist same record for more then one key field.
Can anybody help me to place the logic for how i can check the records with each record? Like
lifnr     banks
1         12
2         34
3         12
4         12
5          76
I want to find out 1,3 and 4.
Thanks a ton in advance

Hi Moni,
IF in my First Reply the Second technique was working then this should solve your Problem:
DATA:
  w_index TYPE i,
  w_lines TYPE i,
  w_duplicate(10) TYPE c.
DESCRIBE itab LINES w_lines.
LOOP AT itab INTO wa.
  w_index = sy-tabix + 1.
  DO. 
  READ TABLE itab INTO wa1 INDEX w_index.
    IF w_index <= w_lines.
      IF wa-banks EQ wa1-banks.
        CONCATENATE w_duplicate wa1-lifnr INTO w_duplicate SEPARATED BY SPACE.
        wa-duplicate = w_duplicate.
        MODIFY itab FROM wa.
      ENDIF.
    ELSE.
      EXIT.
    ENDIF.
   ADD 1 TO w_index.
  ENDDO.
ENDLOOP.
LOOP AT itab INTO wa.
  write:
    / wa-lifnr,
      wa-banks,
      wa-duplicates.
ENDLOOP.
ELSE try the below code it is same as above except one more Internal Table is used:
DATA:
  w_index        TYPE i,
  w_lines        TYPE i,
  jtab            LIKE itab,
  w_duplicate(10) TYPE c.
DESCRIBE itab LINES w_lines.
LOOP AT itab INTO wa.
  w_index = sy-tabix + 1.
  DO. 
  READ TABLE jtab INTO wa1 INDEX w_index.
    IF w_index <= w_lines.
      IF wa-banks EQ wa1-banks.
        CONCATENATE w_duplicate wa1-lifnr INTO w_duplicate SEPARATED BY SPACE.
        wa-duplicate = w_duplicate.
        MODIFY itab FROM wa.
      ENDIF.
    ELSE.
      EXIT.
    ENDIF.
   ADD 1 TO w_index.
  ENDDO.
ENDLOOP.
LOOP AT itab INTO wa.
  write:
    / wa-lifnr,
      wa-banks,
      wa-duplicates.
ENDLOOP.
Regards,
Sunil.

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    curtp = itab2-curtp.
    IF sy-subrc = 0.
    <b>l_sum = sum_pvprs-pvprs + l_sum.</b>
    APPEND sum_pvprs.
    ENDSELECT.
    <b>DESCRIBE TABLE sum_pvprs LINES lrec.
    lavg = l_sum / lrec.</b>
    Message was edited by: Anurag Bankley

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