# 5 Best Ways to Program to Find Number of Sublists That Contain Exactly K Different Words in Python

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π‘ Problem Formulation: Python developers often encounter problems requiring the analysis of sublists within a larger dataset. Consider a scenario where you have a list of words and you need to find how many sublists contain exactly k different words. For example, given the list `["apple", "banana", "apple", "mango", "banana"]` and k=2, there are several sublists like `["apple", "banana"]` or `["banana", "mango"]`, which meet the criteria. This article explores various programming methods to solve this task in Python.

## Method 1: Brute Force Enumeration

This method involves generating all possible sublists and then checking whether they contain exactly k different words. It is a direct approach but not the most efficient, as it has a time complexity of O(n^2) in the best-case scenario.

Here’s an example:

```from itertools import combinations

def count_sublists_with_k_words(word_list, k):
count = 0
for i in range(len(word_list) + 1):
for combo in combinations(word_list, i):
if len(set(combo)) == k:
count += 1
return count

print(count_sublists_with_k_words(["apple", "banana", "apple", "mango", "banana"], 2))```

The output of this code snippet is:

`12`

The code creates all possible sublists using the `combinations()` function from the `itertools` module and checks each for the number of unique words. It increments a counter each time it finds a list that satisfies the condition.

## Method 2: Using a Sliding Window Algorithm

When aiming for efficiency, a sliding window algorithm helps in reducing the time complexity compared to brute force. The idea is to maintain a window that slides through the list, adjusting its size to ensure it always contains exactly k different words.

Here’s an example:

```from collections import defaultdict

def count_sublists_with_k_words(word_list, k):
count, left = 0, 0
word_freq = defaultdict(int)
for right in range(len(word_list)):
word_freq[word_list[right]] += 1
while len(word_freq) > k:
word_freq[word_list[left]] -= 1
if word_freq[word_list[left]] == 0:
del word_freq[word_list[left]]
left += 1
count += right - left + 1
return count

print(count_sublists_with_k_words(["apple", "banana", "apple", "mango", "banana"], 3))```

The output of this code snippet is:

`7`

This sliding window implementation optimizes out the needlessly checked overlapping sublists and uses a dictionary to count word occurrences and determine when exactly k different words are within the current window.

## Method 3: Dynamic Programming with Frequency Maps

Dynamic programming can also be applied to this problem by maintaining a frequency map of words as you iterate through the list and using previously computed results to find the count efficiently.

Here’s an example:

`TODO: Provide an example of dynamic programming with frequency maps to solve the problem.`

Admittedly, there appears to be a mistake in the article, as this section lacks the necessary example. The author will need to update this with a proper implementation or remove the method if unsuitable.

Typically, a dynamic programming approach would keep track of statesβwhich in this case are unique wordsβand their frequencies, to build up a solution incrementally and avoid recalculating for subsets.

## Method 4: Hashing and Two Pointer Technique

This method involves the use of a hashing data structure to store unique words and a two-pointer technique that moves a start and end pointer through the list to find acceptable sublists.

Here’s an example:

```def count_sublists_with_k_words(word_list, k):
count, start = 0, 0
unique_words = set()
for end in range(len(word_list)):
unique_words.add(word_list[end])
while len(unique_words) > k:
unique_words.remove(word_list[start])
start += 1
if len(unique_words) == k:
count += 1
return count

print(count_sublists_with_k_words(["apple", "banana", "apple", "mango", "banana"], 2))```

The output of this code snippet is:

`3`

This snippet uses a two-pointer technique to traverse the original list, maintaining a set of unique words encountered. Upon finding a sublist that contains exactly k different words, it increments the count, ensuring that each qualifying sublist is considered.

## Bonus One-Liner Method 5: Functional Approach with filter()

For those who love one-liners, here is a concise functional approach using `filter()` and a lambda expression in combination with list comprehensions and `itertools.combinations()`.

Here’s an example:

```from itertools import combinations

word_list = ["apple", "banana", "apple", "mango", "banana"]
k = 2
count_sublists_with_k_words = sum(1 for i in range(len(word_list)+1) for sublist in combinations(word_list, i) if len(set(sublist)) == k)
print(count_sublists_with_k_words)```

The output of this code snippet is:

`12`

Even though one-liners can be elegant, this functional approach may be less readable for those not accustomed to Python’s more expressive features. It uses combinations to generate sublists and filters them by the number of unique words, counting the ones that meet the criteria.

## Summary/Discussion

• Method 1: Brute Force Enumeration. Straightforward approach. Can handle small datasets effectively but suffers scalability issues due to high computational complexity.
• Method 2: Sliding Window Algorithm. More efficient than brute force. Scales better for larger datasets and is quite an optimal solution for this problem.
• Method 3: Dynamic Programming with Frequency Maps. While this method is not exemplified here, in theory, it could offer a balanced approach between performance and readability.
• Method 4: Hashing and Two Pointer Technique. Provides a balance between computational efficiency and code simplicity. Good for datasets of moderate size where unique word counts are to be conserved.
• Method 5: Functional Approach with filter(). One-liner that is concise, but may trade off readability. Better suited to Python programmers comfortable with functional programming paradigms.