Finding a String

mystring = "Hello World"
print (mystring.count('ll'))

1

print (mystring.count('l'))

3

obviously, "count" method return the number of a search is found wihtin a string so far as we learned, let's solve a problem from HackerRank at https://www.hackerrank.com/challenges/find-a-string/problem at first, you may think that using "count" is a straight forward solution but soon you realize not to being true.

#input ABCDCDC, CDC
# output: 2
#trying count will return 1
print("ABCDCDC".count('CDC'))

1

Note that we have "CDC" - starting position 2 - starting position 4 - these two results are interlaced. This is the problem - if they are totaly distinct from each other, the solution using 'count' would work perfectly

# have a look at this
print("ABCDCCDC".count('CDC'))

2

#so here is the solution
def count_substring(string, sub_string):
    count = 0
    str_len = len(string)
    sub_len = len(sub_string)

    for x in range (str_len - sub_len + 1):
        scan_string = string[x:x+sub_len]
        if scan_string == sub_string:
            count+= 1

    return count

count_substring('ABCDCCDC','CDC')

2

commenting the solution - we iterate over the string moving 1 position at each cycle then compare - simply our iterating code would be (for x in range (str_len - sub_len + 1) and the results will still work fine but - note that last iterations wont match at all for all scan_string < sub_len - that is why, it is reveal your deep understanding to not enter these loos

mystring ="Hello World"
print( any(char.islower() for char in mystring))

True