Android back end technology language java python ide coderbuddy exo clo

The problem is to find the longest word in a given list of words that starts with a specific prefix. In this case, the prefix is “ex”. We need to write a Python program to solve this problem.

Option 1: Using a for loop

We can solve this problem by iterating through each word in the list and checking if it starts with the given prefix. We can keep track of the longest word found so far and update it whenever we find a longer word.

words = ['Android', 'back', 'end', 'technology', 'language', 'java', 'python', 'ide', 'coderbuddy', 'exo', 'clo']
prefix = 'ex'

longest_word = ''
for word in words:
    if word.startswith(prefix):
        if len(word) > len(longest_word):
            longest_word = word

print("Longest word starting with prefix 'ex':", longest_word)

This solution uses a for loop to iterate through each word in the list. It checks if the word starts with the given prefix using the startswith() method. If it does, it compares the length of the word with the length of the current longest word and updates it if necessary. Finally, it prints the longest word found.

Option 2: Using list comprehension

We can also solve this problem using list comprehension, which is a concise way to create lists in Python. We can create a new list containing only the words that start with the given prefix, and then find the longest word in that list.

words = ['Android', 'back', 'end', 'technology', 'language', 'java', 'python', 'ide', 'coderbuddy', 'exo', 'clo']
prefix = 'ex'

matching_words = [word for word in words if word.startswith(prefix)]
longest_word = max(matching_words, key=len)

print("Longest word starting with prefix 'ex':", longest_word)

This solution uses list comprehension to create a new list matching_words that contains only the words starting with the given prefix. It then finds the longest word in that list using the max() function with the key=len argument, which compares the words based on their lengths. Finally, it prints the longest word found.

Option 3: Using the filter() function

We can also solve this problem using the filter() function, which allows us to filter elements from a list based on a given condition. We can use a lambda function to define the condition, which checks if a word starts with the given prefix. We can then find the longest word in the filtered list.

words = ['Android', 'back', 'end', 'technology', 'language', 'java', 'python', 'ide', 'coderbuddy', 'exo', 'clo']
prefix = 'ex'

matching_words = list(filter(lambda word: word.startswith(prefix), words))
longest_word = max(matching_words, key=len)

print("Longest word starting with prefix 'ex':", longest_word)

This solution uses the filter() function to create a filtered list matching_words that contains only the words starting with the given prefix. It uses a lambda function to define the condition for filtering. It then finds the longest word in that list using the max() function with the key=len argument. Finally, it prints the longest word found.

Among the three options, the best one depends on the specific requirements of the problem and personal preference. Option 1 using a for loop is the most straightforward and easy to understand. Option 2 using list comprehension is more concise and elegant. Option 3 using the filter() function is also concise and allows for more complex filtering conditions. Ultimately, the choice depends on the specific needs of the problem and the coding style of the programmer.

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11 Responses

    1. Hot sauce on pizza? No thanks. Option 3 with filter() might work for some, but its not always the best choice. There are other approaches worth considering too. Its all about finding the right flavor for your code.

  1. I personally prefer Option 3 – the filter() function. Its like finding hidden treasures in your code! 🕵️‍♀️💎

    1. I couldnt disagree more. List comprehension in Python is overrated and often leads to unreadable code. It may seem concise, but it sacrifices clarity. Give me a traditional loop any day. #TeamOption1

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