How to keep count in a recursive function?

Question

I wrote a recursive function to find the number of instances of a substring in the parent string.

The way I am keeping count is by declaring/initialising count

Answer 1

steps = 0
def addPersistence(number, steps):
    steps += 1
    if len(str(number))==1:
        print(str(number) + "\nDone---------------------------------------------------")
        print("TOTAL STEPS " + str(steps-1))

    digits = [int(i) for i in str(number)]

    result = 0
    for j in digits:
        result += j


    if len(str(number)) != 1:
        print(number)
        addPersistence(result, steps)

This is a example copied from one of my projects. This function is to determine to addition persistence of a number (which is adding the digits of a number and repeating it until the number gets to 1) but it surely can be reusable and you can use your function like this (This is Python 3 code, if you want to change it to Python 2, it will still work):

count=0
def countSubStringMatchRecursive(target,key,count):
    index=find(target,key)
    targetstring=target
    if index>=0:
        count+=1
        target=target[index+len(key):]
        countSubStringMatchRecursive(target,key,count)
    else :
        pass
    print("STEPS: "+str(count))

You might notice my code and that code are a little bit different. Why? The answer is that the number of steps it takes for a number to get to 1 using that function is that the last call isn't included in my function because it's a duplication of the call before that.