Need steps for a basic python loop question

# our input variables:
term = 'bob'
s = 'azcbobobegghakl'

# where I store count of found search term:
count = 0

# we are going from 0 to length of input string:
for i in range(len(s)):

    # compare the slice from <current index> to <current index + length of search term> to search term
    if s[i:i+len(term)] == term:
        # it there's match, increase the count
        count += 1

print(count)

Prints:

2

So here is my solution:

s = 'azcbobobegghakl' 
x = 'bob'
count = 0
for i in range(len(s)):
    if x == s[i : i + len(x)]:
        count+= 1
print(count)

Few things to note. First of all your for statement had a couple errors, where you are using the variable count before the statement but also creating a new variable count in the for statement, and so I replaced it in the statement with the variable name i so that it would not override your first count variable. Secondly, it would need to be range(len(s)) as s is a string and you will want to iterate up to it's length.

Finally to explain the if x == s[i : i + len(x)]: part, this is to see if X is the same as the i'th position in S, along with the next 2 letters in S, because the length of X ('bob') is 3, so you'll need to be looking at every len(x) or three letters in the s string.

I did something like this, I am pretty sure someone can optimise the code, I just tried a quick one. It works so far.

keyword = "bob"
text = "azcbobobegghakl"
n = len(keyword)
temp = ""
counter = 0
for x in (text):
    temp += x
    if len(temp)%(n+1) == 0:
        temp = temp[1::]
        if keyword in temp:
            counter += 1
print(counter)

Just a suggestion to complete the answers already given.

You probably could start by checking if the string you are looking for, 'bob', exists in the string to be searched, s, for example:

s = 'azcbobobegghakl'
if 'bob' in s:
    # your searching algotithm

This could also work

s = 'azcbobobegghakl' 
x = 'bob'
count = 0
for i in range(len(s)):
    if s[i:].startswith(x):
            count += 1

I hope this helps ,

Actually this code iterates through the text with the use of while loop

1. text.find(keyword, index) searches for "bob" starting from the current index.

2. If a match is found, count is increased by one.

3. The crucial step is index += 1. This moves the search position forward by just one character, which is why it can find the second "bob" in "azc**bobob**egghakl".

4. The loop stops when .find() can no longer locate the keyword and returns -1.

keyword = "bob"
text = "azcbobobegghakl"
count = 0
index = 0

while index < len(text):
    index = text.find(keyword, index)
    if index == -1:
        break
    count += 1
    index += 1

print(count)