Question

Using Python:

#Write a letterAppearance function that takes a string, str as argument
#letterAppearance will count the number of times each letter appears in a string
#using a dictionary. #letterAppearance returns only the letters that appear in the string.
#Capitalization should not matter
'''
letterAppearance('hello')
#{'h': 1, 'e': 1, 'l': 2, 'o': 1} letterAppearance('Hello')
#{'h': 1, 'e': 1, 'l': 2, 'o': 1} letterAppearance('mutation')
#{'m': 1, 'u': 1, 't': 2, 'a': 1, 'i': 1, 'o': 1, 'n': 1}
'''

Answer 1

The answer to this question is as follows:

The code is as follows:

def letterAppearance(str1):
letter_freq = {}
for i in str1.lower():
#checking the character is already there then incrementing 1
if i in letter_freq:
letter_freq[i] += 1
#otherwise providing the 1
else:
letter_freq[i] = 1
return letter_freq
print(letterAppearance("mutation"))

The input and output is as follows:

1 - def letter Appearance(str1): 2 letter_freq = {} 3- for i in stri.lower(): 4. #checking the character is already there then incrementing 1 5 - if i in letter_freq: 6 letter_freq[i] += 1 7 #otherwise providing the 1 else: letter_freq[i] = return letter_freq 11 print(letterAppearance("mutation")) 12 9 11 10 Execute Mode, Version, Inputs & Arguments 3.7.4 V Stdin Inputs Interactive Command Line Arguments Execute C. Result CPU Time: 0.01 sec(s), Memory: 7796 kilobyte(s) executed in 0.633 sec(s) {'m': 1, 'u': 1, 't': 2, 'a': 1, 'i': 1, 'o': 1, 'n': 1}