Problem Set 1: Counting Bobs

pippy
New Member

Posts: 6

Problem Set 1: Counting Bobs Feb 28, 2014 15:07:32 GMT

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Post by pippy on Feb 28, 2014 15:07:32 GMT

Assume s is a string of lower case characters.

Write a program that prints the number of times the string 'bob' occurs in s.

Is this the correct way to think this through:

bobs = 0
Move through string until you come to a 'b',

if the next character is an 'o',
look at the next character,
if that character is a 'b',
increase 'bobs' by 1
if the next character is not an 'o',
move onto the next character
print 'The number of bobs: %d' % bobs

batmanbury
New Member

Posts: 13

Problem Set 1: Counting Bobs Mar 2, 2014 10:47:11 GMT

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Post by batmanbury on Mar 2, 2014 10:47:11 GMT

Your logic looks sound to me, as long as you're looking at a single character at a time at the beginning.

There are a number of ways to do it. You could also check 3 indexes at a time while looking for 'bob' like this:

[012],[123],[234],[345],[456]...

andreareina New Member Posts: 8	Problem Set 1: Counting Bobs Mar 2, 2014 23:36:00 GMT Quote Select Post Deselect Post Link to Post Member Give Gift Back to Top Post by andreareina on Mar 2, 2014 23:36:00 GMT Remember to account for the case where the last "b" can also be the first of another "bob"

batmanbury
New Member

Posts: 13

Problem Set 1: Counting Bobs Mar 3, 2014 9:05:37 GMT

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Post by batmanbury on Mar 3, 2014 9:05:37 GMT

Mar 2, 2014 23:36:00 GMT andreareina said:

Remember to account for the case where the last "b" can also be the first of another "bob"

Good point. "bobobobob" is actually 4 bobs.

If you're iterating over each character, one by one, that should take care of it naturally.

nick3499
Guest

Problem Set 1: Counting Bobs Jan 11, 2017 18:23:47 GMT

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Post by nick3499 on Jan 11, 2017 18:23:47 GMT

here are a couple of ways to solve that:

s = 'aksjdfhkjadfboboblkajsdhfkljasdkfs'

count = 0
p = 0

while s.find('bob', p) != -1:
    p = s.find('bob', p)
    p += 1
    count += 1

print("Number of times bob occurs is: " + str(count))

s = 'obobcobooloafbobb'
count = 0

for i in range(len(s)):
    if s[i] == 'b' and not i >= (len(s) - 2):
        if s[i + 1] == 'o':
            if s[i + 2] == 'b':
                count += 1

print("Number of times bob occurs is: " + str(count))

Python CS

Problem Set 1: Counting Bobs

Post by pippy on Feb 28, 2014 15:07:32 GMT

Post by batmanbury on Mar 2, 2014 10:47:11 GMT

Post by andreareina on Mar 2, 2014 23:36:00 GMT

Post by batmanbury on Mar 3, 2014 9:05:37 GMT

Post by nick3499 on Jan 11, 2017 18:23:47 GMT

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