5 Best Ways to Count Number of Sublists with Exactly K Unique Elements in Python

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πŸ’‘ Problem Formulation: We often face scenarios where we need to analyze sublists within a larger list, especially to count how many of those sublists contain a specific number of unique elements. Consider we are given a list, say [1, 2, 3, 1, 2], and we want to find the number of sublists that contain exactly k unique elements, for example, if k=2 the desired output should be 7, including sublists like [1,2], [2,3], and [3,1]. This article discusses various methods to effectively program this in Python.

Method 1: Brute Force Approach

This method involves iterating through all possible sublists of the given list and counting those with the desired number of unique elements. It’s the most straightforward approach and serves a good baseline for understanding more complex algorithms. The function typically takes a list and an integer k, and returns the count of sublists with exactly k unique elements.

Here’s an example:

def count_sublists_brute_force(lst, k):
    count = 0
    for i in range(len(lst)):
        for j in range(i+1, len(lst)+1):
            if len(set(lst[i:j])) == k:
                count += 1
    return count

print(count_sublists_brute_force([1, 2, 3, 1, 2], 2))

Output: 7

This code snippet defines a function that checks every possible sublist of the input list for the number of unique elements. If a sublist has k unique elements, the count gets incremented. It finally returns the total count of such sublists. While this method is simple, its time complexity is high; for a list with n elements, it roughly takes O(n^3) operations.

Method 2: Using Combinations from itertools

By utilizing Python’s itertools module, we can generate all the combinations of the list’s indices and then construct sublists accordingly. This approach is more efficient than the brute force method for larger lists, as itertools is optimized for performance. It’s still not the best method for very large lists due to the combinatorial explosion but offers a clear and concise implementation.

Here’s an example:

from itertools import combinations

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

print(count_sublists_combinations([1, 2, 3, 1, 2], 2))

Output: 7

In this code snippet, we generate combinations of elements from the list using the combinations() function and count those with exactly k unique elements. While more efficient than the brute force for most practical case scenarios, the time complexity is still not ideal for very large inputs.

Method 3: Sliding Window Technique

The sliding window technique is highly efficient for this problem as it reduces redundant calculations. The idea is to have a “window” that slides across the list from left to right, checking for the number of unique elements. As the window slides, elements are added or removed from a set, which tracks uniqueness. This method has a better average time complexity than the previous methods.

Here’s an example:

def count_sublists_sliding_window(lst, k):
    count = 0
    for i in range(len(lst)):
        unique_elements = set()
        for j in range(i, len(lst)):
            unique_elements.add(lst[j])
            if len(unique_elements) == k:
                count += 1
            if len(unique_elements) > k:
                break
    return count

print(count_sublists_sliding_window([1, 2, 3, 1, 2], 2))

Output: 7

This code snippet uses the sliding window technique to iterate through the list. For each starting point, a window of unique elements is maintained, and the count is increased each time the window has exactly k unique elements. Once the set has more than k unique elements, the inner loop breaks, saving unnecessary calculations and therefore time.

Method 4: Using Dictionary for Frequency Count

Here, we leverage a dictionary to store the frequency of elements. When we slide through the list, we can efficiently maintain the count of unique elements by increasing or decreasing the dictionary values. This method makes the code more readable and somewhat more efficient by not requiring the costly operation of converting a list slice to a set each time.

Here’s an example:

def count_sublists_frequency_dict(lst, k):
    count = 0
    for i in range(len(lst)):
        freq_dict = {}
        unique_count = 0
        for j in range(i, len(lst)):
            if lst[j] not in freq_dict:
                unique_count += 1
                freq_dict[lst[j]] = 1
            elif freq_dict[lst[j]] == 0:
                unique_count += 1
                freq_dict[lst[j]] += 1
            else:
                freq_dict[lst[j]] += 1

            if unique_count == k:
                count += 1
            elif unique_count > k:
                break
            
    return count

print(count_sublists_frequency_dict([1, 2, 3, 1, 2], 2))

Output: 7

The function count_sublists_frequency_dict iterates through the list and for each index, creates a new dictionary to store the frequency of each element. The frequency dictionary allows us to efficiently track and update the count of unique elements. This strategy is an optimization over a strategy that generates sets because dictionary operations have constant time complexity.

Bonus One-Liner Method 5: Functional Programming Approach

For those who prefer concise code, Python’s functional programming capabilities can be used. Employing itertools.chain and a list comprehension, this one-liner can solve the problem. Caution: this code is less readable and efficient due to its compact nature and is included to demonstrate the wide range of Python’s abilities.

Here’s an example:

from itertools import chain, combinations

def count_sublists_one_liner(lst, k):
    return sum(1 for i in chain.from_iterable(combinations(lst, r) for r in range(len(lst)+1)) if len(set(i)) == k)

print(count_sublists_one_liner([1, 2, 3, 1, 2], 2))

Output: 7

This compact code snippet generates sublists of all possible lengths, calculates the unique elements for each, and sums up those with exactly k unique elements. It leverages the expressiveness of Python’s list comprehensions and itertools module but can be inefficient and somewhat obscure to the reader.

Summary/Discussion

  • Method 1: Brute Force. Simplicity. High time complexity. Best for small lists.
  • Method 2: Using Combinations from itertools. Readable. Still not ideal for large data due to combinatorial growth.
  • Method 3: Sliding Window Technique. Efficient average time complexity. Low redundancy.
  • Method 4: Frequency Dictionary. Readability and Optimized performance. Slightly complex logic.
  • Method 5: Functional Programming. Conciseness. Less readable and efficient. Demonstrates Python’s functional programming capabilities.