Check how many times a number appears within another without strings or lists (Python)

Question

I need to find how many times a number appears within another, for example, if I need to find how many times 22 appears of 2228, I need my output to be 1, or if I want to find 12345 in 1234 it must be 0. I can't use either lists or strings, only if and while.

This is my code so far:

def contarApariciones (numA, numB):
count = 0
while numB != 0:
    digA = numA % 10
    digB = numB % 10
    position = 1
    searchedN = 0
    while searchedN != numA:
        if digB == digA:
            cuenta +=1
            searchedN = digA + position
            posicion *= 10
            numB // 10
            return count
        else: numB // 10

print (contarApariciones(22,2228))
#output 1
print (contarApariciones(12345,1234))
#output 0
print (contarApariciones(808,80808808))
#output 2
print (contarApariciones(1,10111010))
#output 5
print (contarApariciones(0,0))

Answer 1

def contarApariciones(numA, numB):
    if numA == 0 == numB:
        return 1
    count = 0
    mod = 10
    while mod <= numA:
        mod *= 10
    while numB:
        if numB % mod == numA:
            count += 1
            numB //= mod
        else:
            numB //= 10
    return count

For example for 22 and 2228, this first computes mod = 100 to always give us the last two digits of numB. Then:

1. 2228 % 100 is 28. Doesn't match, so drop the 8 and continue with 222.
2. 222 % 100 is 22. Matches, so drop the 22 and continue with 2.
3. 2 % 100 is 2. Doesn't match, so drop the 2 and continue with 0.
4. Actually, don't continue, at 0 we stop.

Passes all your tests as well as these:

for a in range(1000):
    for b in range(10000):
        expect = str(b).count(str(a))
        result = contarApariciones(a, b)
        assert result == expect