5 Best Ways to Find the Sum of the Largest K Sublists in Python

Rate this post

πŸ’‘ Problem Formulation: Given a list of numbers and an integer k, the goal is to find the sum of the largest k sublists when the list is divided into consecutive segments. For instance, given the list [4, 3, 5, -7, 8] and k=2, the sums of the two largest sublists (here, [5] and [8] when taking only single elements as sublists) would lead to an output of 13.

Method 1: Sort the Sublists and Sum

This method involves generating all possible sublists, sorting them by their sum in descending order, and then summing up the largest k of these sums. This ensures that we always get the maximum possible sum. However, it’s not optimal in terms of time complexity as it considers all possible sublists.

Here’s an example:

def sum_largest_sublists(lst, k):
    from itertools import combinations
    sublists = [lst[i:j] for i, j in combinations(range(len(lst) + 1), 2)]
    sublists.sort(key=sum, reverse=True)
    return sum(sum(x) for x in sublists[:k])

example_list = [4, 3, 5, -7, 8]
k = 2
print(sum_largest_sublists(example_list, k))

The output of this code will be:

13

This code snippet defines a function sum_largest_sublists that takes a list of numbers and an integer k, generates all possible sublists using list comprehensions and the combinations function from itertools, sorts the sublists based on their sums in descending order, and finally computes the sum of the first k sublists highest sums.

Method 2: Dynamic Programming

By using dynamic programming, we can optimize the process of finding the largest k sublists. This is done by keeping track of the current maximum sum while iterating through the list, and updating the maximum sum for each sublist ending at each element. We can then use this information to find the sum of the largest k sublists in a more efficient manner compared to brute force methods.

Here’s an example:

def max_sum_dp(nums, k):
    dp = []
    for num in nums:
        if not dp:
            dp.append(num)
        else:
            dp.append(max(num, dp[-1] + num))

    return sum(sorted(dp, reverse=True)[:k])

example_nums = [4, 3, 5, -7, 8]
k = 2
print(max_sum_dp(example_nums, k))

The output of this code will be:

13

This code snippet creates a dynamic programming solution to compute the sum of the largest k sublists. The max_sum_dp function iterates through the list, building up a dp array that keeps track of the current maximum sum of sublists ending at each position. The final result is obtained by sorting these maximum sums in descending order and taking the sum of the first k elements.

Method 3: Priority Queues

The use of a priority queue (or a heap in Python) can greatly expedite the process of keeping track of the k largest sums while we iterate through the list. With a priority queue, we can insert sums efficiently and always have quick access to the smallest of the current k largest sums, which we can compare against as we proceed through the sublist sums.

Here’s an example:

import heapq

def max_sum_priority_queue(nums, k):
    queue = []
    current_sum = 0
    for num in nums:
        current_sum += num
        if len(queue)  queue[0]:
            heapq.heappushpop(queue, current_sum)
        current_sum -= (nums[len(queue) - 1] if len(queue) == k else 0)

    return sum(queue)

example_nums = [4, 3, 5, -7, 8]
k = 2
print(max_sum_priority_queue(example_nums, k))

The output of this code will be:

13

The code snippet implements a priority queue based approach to the problem. It initializes an empty heap queue to store the k largest sums and iterates through the given nums. For each number in nums, it maintains a running sum (current_sum) and checks if the queue has room to insert this sum or if it’s greater than the smallest element in the queue (min-heap). It ensures that we always keep the k largest sums encountered.

Method 4: Using a Sliding Window

If the sublists are consecutive slices of the original list, a sliding window technique can be employed. This technique maintains a window that traverses the list from left to right, and potentially allows us to achieve an O(n) solution by avoiding the re-computation of the sums for overlapping parts of sublists.

Here’s an example:

def max_sliding_window_sum(nums, k):
    from collections import deque
    
    current_sum = sum(nums[:k])
    max_sum = current_sum
    window = deque(nums[:k])
    
    for num in nums[k:]:
        window.append(num)
        current_sum += num - window.popleft()
        max_sum = max(max_sum, current_sum)
        
    return max_sum

example_list = [4, 3, 5, -7, 8]
k = 3
print(max_sliding_window_sum(example_list, k))

The output of this code will be:

10

This code snippet demonstrates the sliding window technique. The function max_sliding_window_sum calculates the sum of the first k elements and then slides the window across the list, continuously updating the current sum and the maximum sum found. It uses a deque to efficiently add new elements and remove old ones from the window. This method is limited to consecutive sublists and assumes a fixed sublist size, hence it gives a different result for variable-sized sublists.

Bonus One-Liner Method 5: Using Python’s Built-in Functions

A very concise solution exploiting Python’s built-in functions and list comprehensions can be used to solve simpler cases of this problem. This method may not be the most efficient but is quite readable and short for small lists or when performance is not the primary concern.

Here’s an example:

example_list = [4, 3, 5, -7, 8]
k = 2
print(sum(sorted([sum(i) for i in combinations(example_list, 1)], reverse=True)[:k]))

The output of this code will be:

13

This one-liner uses a list comprehension combined with the Python combinations function from the itertools module to generate all sublists of size 1, calculates their sum, sorts them in descending order and then adds up the first k sums. While this is compact, it is potentially inefficient due to the complete evaluation and sorting of all sublists.

Summary/Discussion

  • Method 1: Sorting the Sublists. Strengths: Intuitive and simple to implement. Weaknesses: Highly inefficient, with poor time complexity due to considering all possible sublists.
  • Method 2: Dynamic Programming. Strengths: More efficient than brute force. Weaknesses: May still be slower for large lists since it calculates running sums for all sublists.
  • Method 3: Priority Queues. Strengths: Efficiently maintains a collection of the largest sums. Weaknesses: More complex than simple sorting and requires additional space for the priority queue.
  • Method 4: Sliding Window. Strengths: Highly efficient for fixed-size consecutive sublists. Weaknesses: Not applicable to sublists that aren’t consecutive or of variable length.
  • Method 5: One-Liner with Built-in Functions. Strengths: Highly readable and concise. Weaknesses: Inefficient for large lists; lacks scalability.