Mantain a count of subarrays that satisfy the condition usning recursion

Question

So I made a recursive function that gives me all the subarrays , I want to apply a condition on those sub-arrays and then keep a count of subarrays that satisfy the condition. But Initialization of count variable has been bothering me ,please help!

here is my code:

def printSubArrays(arr, start, end):
 if end == len(arr):
     return
 elif start > end:
     return printSubArrays(arr,0,end+1)

 else:
     dictio = arr[start:end + 1]
     print(dictio)
     if len(dictio)!=1:
         for i in range(len(dictio)):
             aand =1
             aand =aand & dictio[i]
         if aand %2 !=0:
             count=count+1
     return printSubArrays(arr,start+1,end)

arr=[1,2,5,11,15]
dictio=[]
count = 0
printSubArrays(arr,0,0)
print(count)

Answer 1

The most common technique to keep a count is to use a helper function. So you'd have your principal function call a helper like this:

def printSubArrays(arr, start, end):
   return _printSubArrays(arr, start, end, 0)

The 0 at the end is the count.

Then each time you recurse you increment:

def _printSubArrays(arr, start, end, count):
 if end == len(arr):
     return count
 elif start > end:
     return _printSubArrays(arr,0,end+1, count + 1)

 else:
     dictio = arr[start:end + 1]
     print(dictio)
     if len(dictio)!=1:
         for i in range(len(dictio)):
             aand =1
             aand =aand & dictio[i]
         if aand %2 !=0:
             count=count+1
     return _printSubArrays(arr,start+1,end, count+1)