5 Best Ways to Find Sublists with High Averages in Python

Rate this post

πŸ’‘ Problem Formulation: We seek efficient Python solutions to find all sublists of a given length k within a list, such that the average of the elements in each sublist is greater than or equal to a specified target value. For instance, given a list [1, 2, 3, 4, 5], length k=3, and target average 3, the program should return the count of qualifying sublists.

Method 1: Brute Force

The brute force approach involves generating all possible sublists of length k and then calculating the average of each sublist to check against the target average. Though straightforward, this method’s time complexity can be high for large lists due to the number of sublists to consider.

Here’s an example:

def count_sublists_brute_force(lst, k, target):
    count = 0
    for i in range(len(lst) - k + 1):
        if sum(lst[i:i+k]) / k >= target:
            count += 1
    return count

# Example usage
print(count_sublists_brute_force([1, 2, 3, 4, 5], 3, 3))

Output:

3

This function iterates through the list, creating sublists of size k starting from each index and computes their averages. It counts the sublists that satisfy the condition of having an average greater or equal to the target.

Method 2: Sliding Window

The sliding window technique improves efficiency by avoiding recalculation of the sum for overlapping parts of sublists. As the window slides over the list, it adjusts the sum by adding the incoming element and subtracting the outgoing one, maintaining a constant time operation.

Here’s an example:

def count_sublists_sliding_window(lst, k, target):
    count = 0
    window_sum = sum(lst[:k])
    if window_sum / k >= target:
        count += 1
    for i in range(k, len(lst)):
        window_sum += lst[i] - lst[i - k]
        if window_sum / k >= target:
            count += 1
    return count

# Example usage
print(count_sublists_sliding_window([1, 2, 3, 4, 5], 3, 3))

Output:

3

This function uses a sliding window to keep track of the sum of the current sublist of size k. It efficiently computes the average by updating the sum when the window is moved, increasing performance on large lists.

Method 3: Prefix Sum Array

Prefix sum arrays provide another method to compute sublist averages quickly. By precomputing cumulative sums, we can find the sum of any sublist in constant time. This method also scales better with large lists and can be more efficient than brute force.

Here’s an example:

def count_sublists_prefix_sum(lst, k, target):
    count = 0
    prefix_sum = [0]
    for num in lst:
        prefix_sum.append(prefix_sum[-1] + num)
    for i in range(k, len(prefix_sum)):
        if (prefix_sum[i] - prefix_sum[i - k]) / k >= target:
            count += 1
    return count

# Example usage
print(count_sublists_prefix_sum([1, 2, 3, 4, 5], 3, 3))

Output:

3

The function first constructs a prefix sum array and then uses it to compute the sum of sublists of length k in constant time, thus efficiently addressing the problem and minimizing redundant calculations.

Method 4: Optimized Prefix Sum with Early Stopping

Enhancing the prefix sum method, we can implement early stopping to reduce calculations further if the remaining elements cannot possibly satisfy the average criteria. This method is particularly efficient if we have a sorted list or if we can early predict the impossibility of meeting the target.

Here’s an example:

def count_sublists_optimized_prefix(lst, k, target):
    count = 0
    prefix_sum = [0]
    required_sum = k * target
    for num in lst:
        prefix_sum.append(prefix_sum[-1] + num)
    for i in range(k, len(prefix_sum)):
        if prefix_sum[-1] - prefix_sum[i - k] = required_sum:
            count += 1
    return count

# Example usage
print(count_sublists_optimized_prefix([1, 2, 3, 4, 5], 3, 3))

Output:

3

Similar to the previous prefix sum method, this one incorporates an early stopping condition to cease computations when the remaining elements of the list are inferior to the sum needed to reach the target average.

Bonus One-Liner Method 5: List Comprehensions

List comprehensions in Python provide a concise and readable way to filter sublists. This strategy combines readability with the efficiency of a brute-force method but may not be the fastest approach for large data sets.

Here’s an example:

count_sublists_list_comp = lambda lst, k, target: sum(1 for i in range(len(lst) - k + 1) if sum(lst[i:i+k]) / k >= target)

# Example usage
print(count_sublists_list_comp([1, 2, 3, 4, 5], 3, 3))

Output:

3

This one-liner uses a lambda function with a list comprehension to count sublist averages. It succinctly reproduces the brute-force logic in a single, functional line of code.

Summary/Discussion

  • Method 1: Brute Force. Easy-to-understand and implement. Poor efficiency for large lists.
  • Method 2: Sliding Window. Significantly more efficient than brute force, especially for large lists with minimal additional memory usage.
  • Method 3: Prefix Sum Array. Offers efficient computation of sublists sums. Requires additional memory for the prefix sum array but scales well with list size.
  • Method 4: Optimized Prefix Sum with Early Stopping. Introduces early termination for efficiency. Particularly useful when higher average values are rare or the list is sorted.
  • Method 5: List Comprehensions. Aesthetically pleasing and Pythonic. Best for smaller lists or when code brevity is a priority over performance.