Python Hello I am trying to learn basic recursive function on Python and I am do

Question

Python

Hello I am trying to learn basic recursive function on Python and I am doing some of few exercises for practice. I am trying to make a function that returns the number that numbers in the list are strictly increasing using recursive, but counting algorithm is problem here is my code so far...

def count_increasing_number(string, number=0):
   #If string lenth is 1, answer is 1
   counter = 1
      
   #Base condition
   if(number > int(string)):
       return 0

   for i in range(1, len(string)):
       #If first part is greater than previous number
      
       if int(string[:i]) > number:
      
           #If last part is empty, increment ans
           res_string = string[i:]
           if len(res_string) == 0:
               counter += 1
              
           #Proceed with the remaining part
           else:
               number = int(string[:i])
               counter += count_increasing_number(res_string, number)
   return counter

print(count_increasing_number('123'))
print(count_increasing_number('19325'))
print(count_increasing_number('1111'))
print(count_increasing_number('10203'))


I want to get

3
4
2
4

which are
[1,2,3],[1,23], [123]
[1,9, 325], [1,9325], [19, 325], [19325]
[1,111], [1111]
[1, 02, 03], [1, 0203], [10, 203], [10203]

but the output of my code are

3
4
3
4

There seems to be a problem with the counting algorithm, but I don't know where it's wrong.
If possible, could you let it print a list of numbers as well as counter values?

for example)
[1,2,3],[1,23], [123]
[1,9, 325], [1,9325], [19, 325], [19325]
[1,111], [1111], [????]
[1, 02, 03], [1, 0203], [10, 203], [10203]

I want to find out the error and correct it, but I don't know how to get a list of numbers out, so it's hard to find out what went wrong. I know this is really stupid question, but I cannot find the solution. Please help me....

Explanation / Answer

The code is perfect except one minor change. I have corrected it.

#Base condition
   if(number >=int(string)):
       return 0

It was before "if(number >int(string)):"

Because number = 0 when it call function count_increasing_number() and int(string) will never be less than zero. So I put '>=' which gives correct result.

The corrected code:

def count_increasing_number(string, number=0):
   #If string lenth is 1, answer is 1
   counter = 1

   #Base condition
   if(number >= int(string)):        # Modification done here
       return 0

   for i in range(1, len(string)):
       #If first part is greater than previous number

       if int(string[:i]) > number:
#          print("string",int(string[:i]))    
           #If last part is empty, increment ans
           res_string = string[i:]
#           print("res_string",res_string)
           if len(res_string) == 0:
               counter += 1

           #Proceed with the remaining part
           else:
               number = int(string[:i])
               counter += count_increasing_number(res_string, number)
#              print("else part",counter)
   return counter

print(count_increasing_number('123'))
print(count_increasing_number('19325'))
print(count_increasing_number('1111'))
print(count_increasing_number('10203'))

The output is:

3
4
2
4

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.