1.find most occurrences of a character.
for exm: input:aaabbbbdddddyyy
output: d 5 times
how can i get that?
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Answer / manish dwivedi
By using SCAN OP- Code we can find out the most occurrences
of a character.
| Is This Answer Correct ? | 5 Yes | 1 No |
Answer / kamlesh
you need to write dow loop ,initialize counter variable=0
and use scan opcode
when first occur of 'd' finds scan returns its position
in string , increment counter=counter+1 then you have to
use substr(yrstring:pos+1:'d')now again same process
is done and if 'd' char found increment counter by 1
this process is continue until complete yrstring is scaned.
finally total number of 'd' is accumulate in COUNTER
variable.
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / bharatiya
I was trying it last night and wrote the below code that works for this scenario, please have a look----
DArr1 S 1 DIM(10)
DArr2 S 1 DIM(10)
DCnt S 2 0
Dn1 S 2 0 Inz(1)
Dn2 S 2 0 Inz(1)
DFld1 S 15 Inz('aaabbbbdddddyyy')
C
C Do 15
C EVAL Arr1(n1) = %subst(FLD1:n1:1)
C If Arr1(n1) = 'd'
C Eval Arr2(n2) = Arr1(n1)
C Eval n2 = n2 +1
C Endif
C Eval n1 = n1+1
C Enddo
C Eval cnt = n2 - 1
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / kar
Use Scan Opcode, and give the result field as array name
instead of a ordinary field.
After the execution of statement, display array values, and
you can find the all occurrences.
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / ssss
fld1 = aaabbbbdddddyyy
res = %check('d':fld1:8) ;
If res > 0;
Cnt = res-8;
Endif;
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / vipul dixit
With the help of %Scan built in function or Scan op-code we
can do this.
we should define a data structure and write like this way:-
DDs s
C Eval Ds = %Scan('d':Source:1).
It will fetch the position of the scanned character,and
after that we will count all the positions as DS elements.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / manoj
As per my understanding using Scan opcode we can't find out
most occurrences of a character.
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / deepak
D:a 15a inz('aaabbbbdddddyyy')
D:b 1a
D:c 15a
D:i 2p 0
D:e 2p 0 inz(1)
C: Eval b=%sst(a:e:1)
C: eval c *cat b
C: If b='d'
C: eval replace(d:x)
C: eval i=i+1
C: eval e=e+1
C: i dsply
C: seton lr
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / prasanna
You can find it with opcode scan and in the result field we
should have an array
| Is This Answer Correct ? | 0 Yes | 2 No |
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