5 Best Ways to Find the Longest Contiguous Sublist of Same First Letter Words in Python

💡 Problem Formulation: You are given a list of words and your task is to write a program to find the length of the longest contiguous sublist in which all words start with the same letter. For instance, given the list [“alpha”, “ape”, “baby”, “bear”, “dog”, “duck”], the desired output is 2, representing the two words “alpha” and “ape” that start with the letter ‘a’.

Method 1: Iterative Approach with For-Loop

This method involves iterating through the list of words using a for-loop, tracking the current first letter and the length of the longest sublist thus far. It is simple and intuitive, relying on basic control structures in Python.

Here’s an example:

def longest_sublist(words):
    max_length = 0
    current_length = 0
    prev_letter = ''
    for word in words:
        if word and (word[0] == prev_letter or prev_letter == ''):
            current_length += 1
        else:
            max_length = max(max_length, current_length)
            current_length = 1
        prev_letter = word[0]
    return max(max_length, current_length)

words = ["alpha", "ape", "baby", "bear", "dog", "duck"]
print(longest_sublist(words))

Output:

The longest_sublist function iterates over the list of words, maintaining a count of the length of the longest sublist found with the same initial character. It resets the count when a word starting with a different letter is encountered, and it updates the maximum length found after examining each word.

Method 2: Using itertools.groupby

This strategy utilizes the convenience of Python’s itertools.groupby function, which groups words by their first letter and avoids explicit iteration and state management in the code.

Here’s an example:

from itertools import groupby

def longest_sublist(words):
    return max(len(list(group)) for _, group in groupby(words, key=lambda x: x[0]))

words = ["alpha", "ape", "baby", "bear", "dog", "duck"]
print(longest_sublist(words))

Output:

The longest_sublist function utilizes groupby to cluster the words based on the first character. Then, it calculates the maximum length of these grouped sublists to find the length of the longest contiguous sublist of words with the same initial letter.

Method 3: List Comprehension and zip

This method leverages a list comprehension in conjunction with the zip function to compare adjacent items in the list, allowing for a concise expression of the problem.

Here’s an example:

def longest_sublist(words):
    groups = [1] + [1 if words[i][0] != words[i-1][0] else 0 for i in range(1, len(words))]
    return max(sum(group) for is_new, group in groupby(groups) if not is_new)

words = ["alpha", "ape", "baby", "bear", "dog", "duck"]
print(longest_sublist(words))

Output:

The code snippet defines longest_sublist which computes a list of 1’s and 0’s representing when words start with a new first letter. The groupby function is used again to find contiguous sublists with the same letter, summing the groups of 0’s which represent words with the same initial letter.

Method 4: Enumerate and Max-Heap

In this method, enumerate is paired with a max-heap data structure to keep track of the longest contiguous sublist lengths, efficiently maintaining the maximum length encountered with the help of the heap.

Here’s an example:

import heapq

def longest_sublist(words):
    heap = []
    prev_letter = ''
    count = 0
    for _, word in enumerate(words):
        if word and (word[0] == prev_letter or prev_letter == ''):
            count += 1
        else:
            heapq.heappush(heap, -count)
            count = 1
        prev_letter = word[0]
    heapq.heappush(heap, -count)
    return -heap[0]

words = ["alpha", "ape", "baby", "bear", "dog", "duck"]
print(longest_sublist(words))

Output:

The longest_sublist function creates a max-heap (simulated with negative values) to track the maximum sublist length. The use of enumerate is superfluous here and is shown for illustrative purposes, adding complexity with little benefit. Still, it represents a more algorithmically-rich approach to the problem.

Bonus One-Liner Method 5: Using max with map and groupby

This is a compact one-liner utilizing Python’s powerful built-in functions to find the maximum group length, showcasing how concise Python code can be when taking advantage of its functional aspects.

Here’s an example:

from itertools import groupby

words = ["alpha", "ape", "baby", "bear", "dog", "duck"]
print(max(map(lambda x: len(list(x[1])), groupby(words, key=lambda w: w[0]))))

Output:

The one-liner code snippet applies groupby to the list of words and then uses map to transform each group into its length. The max function then simply picks the largest of these lengths. This solution is elegant but potentially sacrifices some readability for brevity.

Summary/Discussion

• Method 1: Iterative Approach with For-Loop. It’s simple and intuitive. The drawback is manually tracking the current group and maximum found, which can be error-prone.
• Method 2: Using itertools.groupby. This method is clean and relies on Python’s standard library to abstract away the grouping mechanic. However, it can be less efficient with memory as it creates temporary lists.
• Method 3: List Comprehension and zip. This method uses more advanced Python features for a compact solution, but can be more difficult to understand for beginners.
• Method 4: Enumerate and Max-Heap. Offers an algorithmic approach with a performance edge when dealing with extremely large lists. However, it is more complex and somewhat over-engineered for the task.
• Method 5: Bonus One-Liner Using max with map and groupby. It is terse and ‘Pythonic,’ albeit potentially cryptic for those not already familiar with functional programming idioms in Python.