Python | Isogram Problem!

Hey everone, I wanna share my last challenge that I solved, it took me around 2 days!
I know it could be easy for some of you even for beginners but it doesn’t matter for me cause my goal is improving my coding skill and problem solving every single day.
Long story short!
Let’s dive to the Isogram Challenge!

SOLVE IT!


Problem:

Determine if a word or phrase is an isogram.

An isogram (also known as a "nonpattern word"),
 is a word or phrase without a repeating letter,
 however spaces and hyphens are allowed to appear
 multiple times.

Examples of isograms:

- lumberjacks
- background
- downstream
- six-year-old

The word **isograms**,
is not an isogram, because the s repeats.

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Let’s break down the problem to the 3 steps:

1. Making a function
2. We need to use if statement to count if there’s more than one item in our string!
3. we need to loop over if statement to check every item in our string!.

1.Making a function:

  def isogram(string):
       pass

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2.if statement to count if there’s one more item or not:

if string.count(item) > 1 :
            return False

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3.for loop over if statement to check every item in given string:

for item in upper_string:

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*after this steps we need to combine number 2 and number 3, like this:

for item in upper_string:
        if upper_string.count(item) > 1 :
            return False

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So we have a function:

def isogram(string):

    for item in string:
        if string.count(item) > 1 :
            return False
    return True

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If we test our program with a string like this:

abc = isogram("A B C")

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It will return False !
Cause our program count spaces so we have two spaces in “A B C” string!
If your read the Problem part carefully it says:

spaces and hyphens are allowed to appear
multiple times!
So:

spaces = " "
hyphens = "-"

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We need another step to complete our solution which is:
.
.
.
4.writing a __if statement_ for spaces and hyphens to return True if there’s more than once_:

spaces = " "
hyphens = "-"
if string.count(spaces)>1 or string.count(hyphens)>1:
      return True

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and our string() function will be look like this:

def isogram(string):

    for item in string:
        spaces = " "
        hyphens = "-"
        if  string.count(item) > 1 :
            return False
        elif string.count(spaces) > 1 or string.count(hyphens) > 1:
            return True 
        else:
            pass
    return True

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Let’s test our program with another string:

abc = isogram("A a B C")
print(abc)

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Our program will return True!!!
But we have two letter “A” and “a”!
That’s obvious cause there is a difference between Capital letters and small letter in every programming language!
and if you try :

print(A is a)

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the answer would be False.
So in our program we can make all letters Upper-Case(upper()) or Lower-Case(lower()), it’s totally up to you, I prefer upper!

def is_isogram(string):

    upper_string = string.upper()

    for item in upper_string:
        spaces = " "
        hyphens = "-"
        if upper_string.count(item) > 1 :
            return False

        elif string.count(spaces) > 1 or string.count(hyphens) > 1:
            return True 
        else:
            pass
    return True

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I need to say thank you my brother Amir, cause you’re helping me a lot in this way.

Finally we completed this challenge, if you have any suggestion to improve this solution and write more Pythonic, cleaner and more readable just let me know and leave a comment below my post, I’ll appreciate it.
Thank you so much for reading my post.

Keep Moving Forward ツ

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原文链接：Python | Isogram Problem!