Printing Non Repeating Characters first - oracle

Question

I am new to PL/SQL and I am trying to write a procedure which would print the non repeating characters of an input string first and Repeating characters of the string in the last. For example if the input string is "Array" then the output should be "yArra".
I wrote a part of it for searching the no. of occurrences of a repeating character, but don't know how exactly should it be printed at the first place.
I wrote an algorithm on this on how can this be made to work but finding difficult to code
Thanks in advance for your help!

Answer 1

I am trying to write a procedure which would print the non repeating
characters of an input string first and Repeating characters of the
string in the last.
You can do this using a pure PLSQL code as below:
create or replace procedure prnt_letter(strng varchar2) as
var varchar2(1);
var1 varchar2(1000) := '';
var2 varchar2(1);
var3 varchar2(1000) := '';
strn_len number;
begin
dbms_output.put_line('Input String --> ' || strng);
strn_len := length(strng);
var := substr(strng, 1, 1);
for i in 1 .. strn_len loop
if (var = substr(strng, i, 1)) then
var2 := substr(strng, i, 1);
var3 := var3 || var2;
var := substr(strng, i, 1);
else
var1 := var1 || substr(strng, i, 1);
var := substr(strng, i, 1);
end if;
end loop;
dbms_output.put_line('Output String --> '||var1 || var3);
end;
EDIT:
Here is my revised solution both in PLSQL and SQL. This works for any string.
PLSQL:
create or replace procedure prnt_letter(strng varchar2) as
var1 varchar2(1000) := '';
strn_len number;
begin
dbms_output.put_line('Input String --> ' || strng);
strn_len := length(strng);
SELECT reverse (LISTAGG (vertical, '') WITHIN GROUP (ORDER BY 1 DESC))
into var1
FROM (
SELECT SUBSTR (strng, LEVEL, 1) Vertical
FROM DUAL
CONNECT BY LEVEL <= strn_len
) ;
dbms_output.put_line('Output String --> '||var1 );
end;
Output:
SQL> execute prnt_letter('rajjjjkkmmaaljjjl');
Input String --> rajjjjkkmmaaljjjl
Output String --> rmmllkkjjjjjjjaaa
PL/SQL procedure successfully completed.
SQL> execute prnt_letter('bubble');
Input String --> bubble
Output String --> ulebbb
PL/SQL procedure successfully completed.
SQL:
-- Logic used:
1) The input string is first arranged vertically in separate rows and
then ordered
2) Using LISTAGG, the result was assembled as a single ordered string
3) Using REVERSE the non-repeating string is brought to the starting
of the string.
SELECT reverse (LISTAGG (vertical, '') WITHIN GROUP (ORDER BY 1 DESC)) col1
FROM ( SELECT SUBSTR ('rajjjjkkmmaaljjjl', LEVEL, 1) Vertical
FROM DUAL
CONNECT BY LEVEL <= LENGTH ('rajjjjkkmmaaljjjl')
)

Answer 2

Try and use the REGEXP_COUNT function for the same. You can first provide a filter where this result >1 to find repeating characters and then concatenate them with the ones whose count = 1.
Check how to use regexp_count

Answer 3

I think the solution can be acheived by just using pure SQL rather than using PLSQL. Hope below snippet helps.
SELECT a.COL
||REPLACE('&Enter_text',a.col,'') output
FROM
(SELECT regexp_count('&Enter_text',SUBSTR('&Enter_text',level,1)) col1,
SUBSTR('&Enter_text',level,1) col
FROM DUAL
CONNECT BY level <=LENGTH('&Enter_text')
)a
WHERE a.col1 = 1;

Answer 4

That's fun, so I came out with something easily understandable using associative arrays as Hashmap; there's something subtle also with the non-case-sensitiveness:
CREATE OR REPLACE FUNCTION f(p_str in varchar2)
RETURN varchar2
AS
TYPE map_v IS TABLE OF integer INDEX BY varchar2(1);
l_dup map_v;
i PLS_INTEGER;
l_c varchar2(1);
l_tc varchar2(1);
l_nb_occurrences integer := NULL;
l_out_sngl varchar2(2000) := '';
l_out_dupl varchar2(2000) := '';
BEGIN
-- l_dup('a'):=0;
-- l_dup('b'):=0;
-- first loop to count occurrences
i:=1;
LOOP
l_c := lower(substr(p_str, i, 1));
begin
l_nb_occurrences := l_dup(l_c);
l_dup(l_c) := l_nb_occurrences + 1;
dbms_output.put_line(l_c||':incr:'||i);
exception
when no_data_found then
l_dup(l_c) := 1;
dbms_output.put_line(l_c||':pushed:'||i);
when others then
raise;
end;
i := i+1;
EXIT WHEN i > length(p_str);
END LOOP;
-- second loop for building output
i:=1;
LOOP
l_c := lower(substr(p_str, i, 1));
l_tc := substr(p_str, i, 1);
begin
l_nb_occurrences := l_dup(l_c);
dbms_output.put_line(l_c||':xx:'||i||'||'||l_nb_occurrences);
if l_nb_occurrences = 1 then
l_out_sngl := l_out_sngl || l_tc;
else
l_out_dupl := l_out_dupl || l_tc;
end if;
exception
when no_data_found then
dbms_output.put_line('why? there should be (see first loop).');
when others then
raise;
end;
i := i+1;
EXIT WHEN i > length(p_str);
END LOOP;
return l_out_sngl || l_out_dupl;
exception
when others then
dbms_output.put_line(sqlerrm);
END f;
/
Which gives results:
select f('Array') from dual;
-- yArra
select f('Bubbles') from dual;
-- ulesBbb

Answer 5

Try this function...
CREATE OR REPLACE FUNCTION non_repeating_char_first (v_input IN varchar2)
RETURN varchar2
IS
str1 varchar2 (100);
str2 varchar2 (100);
v_match number;
v_output varchar2 (100);
BEGIN
str1 := '';
str2 := '';
FOR i IN 1 .. LENGTH (v_input)
LOOP
SELECT REGEXP_COUNT (v_input,
SUBSTR (v_input, i, 1),
1,
'i')
INTO v_match
FROM DUAL;
IF v_match =1 THEN
str1 :=str1||SUBSTR (v_input, i, 1);
else
str2 :=str2||SUBSTR (v_input, i, 1);
END IF;
END LOOP;
v_output:=str1||str2;
RETURN v_output;
END;