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Count all prefixes in given string with greatest frequency using Python
In this tutorial, we are going to write a program that counts and prints the words with a higher frequency of an alphabet than the second one.
Take a string and two alphabets. The prefixes with a higher frequency of the first alphabet will be printed. And display the count at the end of the output.
Let's see some examples.
Input
string:- apple alphabets:- p, e
Output
ap app appl apple 4
Input
string:- apple alphabets:- e, p
Output
0
Let's see the steps to write the code.
Define a function and write the code in it.
Initialize count to 0 and an empty string.
Iterate over the string.
Get the prefix using the string slicing and index. And store it in an empty string.
Compare the alphabet frequency in the prefix.
Print and increment the count if satisfied.
Print the count at the end.
Example
# defining a function for multiple calles
def prefixes(string, _1, _2):
# count count = 0
# empty string for comparison
prefix = ""
# iterating over the string
for i in range(len(string)):
# getting the prefix from the string
prefix = string[:i + 1]
# comparing the count of alphabets in the prefix
if prefix.count(_1) > prefix.count(_2):
# printing the prefix if success
print(prefix)
# incrementing the count by 1
count += 1
# printing the count
print(f"Total prefixes matched: {count}")
if __name__ == '__main__':
# invokging the function
print(f"----------------apple p e---------------------")
prefixes('apple', 'p', 'e')
print()
print(f"----------------apple e p---------------------")
prefixes('apple', 'e', 'p')Output
If you run the above code, you will get the following result.
----------------apple p e--------------------- ap app appl apple Total prefixes matched: 4 ----------------apple e p--------------------- Total prefixes matched: 0
Conclusion
If you are facing any problems in understanding the code, mention them in the comment section.
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