Practice Problems: Not understanding the Auxiliary function?

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#1

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In this practice problem I don’t understand the solution to the part dealing with the auxiliary function. The aim was to create a frequency table of the characters in a string. I have commented on what is confusing me below.

Code: 

def frequency(string):
    freq = {}
    for c in string:
        if c not in freq:
            freq[c] = 0  # Why is this not freq[c] = 1
        freq[c] += 1
    return freq
frequency('aaabbc') #Why does this not return {a:2 , b: 1, c:0} based on 
                    #line 5 of this block of code.
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#2

Because in the next line it is going to add by one either way if condition is True or False
here

I think you are confusing given solution with this logic.

def frequency(string):
    freq = {}
    for c in string:
        if c not in freq:
            freq[c] = 1
        else:
            freq[c] += 1
    return freq
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#3

So will this code run irrespective of the result of the result of if c not in freq:? If so why even include an if-statement?

#4

First time you need to add new character in the dictionary if not available in the dictionary. else you will get runtime error.

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#5

So could you use the following code and get the same result?

#6

Yes, You can try it. you will get same results.

1 Like
#7

Thanks so much for your help!

1 Like
#8

Hello
even with the solution provided i am still facing problem. can you help?

def character_freq(string):
    freq = {}
    for c in string:
        if c not in freq:
            freq[c] = 0
        else:  
            freq[c] += 1
    return freq
def are_anagrams(string1, string2):
    if len(string1) != len(string2):
        return False
    freq1 = character_freq(string1)
    freq2 = character_freq(string2)
    for c in freq1:
        if c not in freq2 or freq1[c] != freq2[c]:
            return False
        return True
print(are_anagrams('gainly', 'laying'))
print(are_anagrams('banana', 'bacana'))
#9

Hi, what problem are you facing?

#10

Hello,
The output is supposed to be:
True
False
And also I got this message from the system when i am trying to submit:
Function are_anagrams did not return the expected value.
With this code provided my output is
True
True
So something is wrong
Thank you

#11

Problem seems to indentation (the return statement) in are_anagrams function. It should be

def are_anagrams(string1, string2):
    if len(string1) != len(string2):
        return False
    freq1 = character_freq(string1)
    freq2 = character_freq(string2)
    for c in freq1:
        if c not in freq2 or freq1[c] != freq2[c]:
            return False
    return True
1 Like
#12

Perfect! Thank you very much! :grinning:

1 Like
#13

I have another way of solve, it’s about using set element to check whether if two strings are anagrams of each other
Look at these examples:

test1_string1 = 'gainly'
test1_string2 = 'laying'
# Expected output: True

test2_string1 = 'banana'
test2_string2 = 'bacana'
# Expected output: False

We can see that whatever kind of re-arrange position element in a string is, if one is an anagram, the element in the string isn’t replaced by any different element, it means that the exists of element in string is the same but only position change

So I use set condition to check that, here is my code:

def are_anagrams(string1, string2):
    set_1 = set(string1)
    set_2 = set(string2)
    if set_1 == set_2:
        return True
    else:
        return False

I dunno why, because while we can go with a easier way, did we have any edge case that make this function is error? Because I check with the provided input and the result is OK