5 Best Ways to Find the Sum of All Contiguous Sublists in Python

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πŸ’‘ Problem Formulation: We aim to compute the cumulative sum of all possible contiguous subarrays within a given list of numbers. For instance, given the list [1, 2, 3], the desired output would be the sum of [1], [2], [3], [1, 2], [2, 3], and [1, 2, 3], which equals to 20.

Method 1: Brute Force Approach

This method involves using a triple nested loop to consider every possible contiguous sublist and calculate their sums. Although this approach is straightforward and easy to understand, it has a considerable drawback in terms of efficiency due to its O(n^3) time complexity, where n is the length of the original list.

Here’s an example:

def sum_of_sublists(lst):
    total = 0
    for i in range(len(lst)):
        for j in range(i, len(lst)):
            total += sum(lst[i:j+1])
    return total

# Example list
example_list = [1, 2, 3]
print(sum_of_sublists(example_list))

Output: 20

The provided function sum_of_sublists() iterates through all the possible starting and ending points of sublists within a given list, calculates the sum of each sublist, and adds it to the total sum. It returns the cumulative sum after considering all sublists.

Method 2: Prefix Sum Technique

The prefix sum technique improves the efficiency by pre-calculating the cumulative sum at each index, reducing the complexity of the solution to O(n^2). By avoiding the repeated summation of elements each time a new sublist sum is calculated, this method provides a significant speed-up over the brute force approach.

Here’s an example:

def sum_of_sublists(lst):
    prefix_sums = [0] * (len(lst) + 1)
    for i in range(len(lst)):
        prefix_sums[i+1] = prefix_sums[i] + lst[i]
    
    total = 0
    for i in range(len(lst)):
        for j in range(i, len(lst)):
            total += prefix_sums[j+1] - prefix_sums[i]
    return total

# Example list
example_list = [1, 2, 3]
print(sum_of_sublists(example_list))

Output: 20

The function sum_of_sublists() uses an additional list to store prefix sums, which are then used to quickly calculate the sum of sublists between indices i and j. The efficiency gain is achieved through the reduction of redundant computations by direct subtraction.

Method 3: Cumulative Index Method

This method leverages the contribution of each element to the final sum, which depends on its position within the array. By multiplying the value at index i by the number of times it will appear in the sublists, we can obtain the sum in O(n) time.

Here’s an example:

def sum_of_sublists(lst):
    total = 0
    n = len(lst)
    for i, num in enumerate(lst):
        total += num * (i + 1) * (n - i)
    return total

# Example list
example_list = [1, 2, 3]
print(sum_of_sublists(example_list))

Output: 20

The sum_of_sublists() function calculates the total sum by iterating over the list once and considering the number of sublists each element belongs to, which is determined by its position in the list. This results in a much more efficient calculation.

Method 4: Dynamic Programming

Dynamic programming can also be utilized for this problem by constructing a solution based on the sums of previously computed sublists. The idea is to build up the solution using the results of smaller sub-problems in order to achieve an O(n^2) time complexity.

Here’s an example:

def sum_of_sublists(lst):
    n = len(lst)
    sum_matrix = [[0] * n for _ in range(n)]
    total = 0

    for i in range(n):
        for j in range(i, n):
            sum_matrix[i][j] = lst[j] + (sum_matrix[i][j-1] if j > i else 0)
            total += sum_matrix[i][j]

    return total

# Example list
example_list = [1, 2, 3]
print(sum_of_sublists(example_list))

Output: 20

By leveraging a matrix to store intermediate sums of sublists, the sum_of_sublists() function calculates the total in a bottom-up manner, reusing previous results to avoid redundant computations.

Bonus One-Liner Method 5: List Comprehension

For those who prefer compact code, Python’s list comprehensions can be used for an elegant one-liner solution. This method trades off readability for brevity and maintains O(n^3) complexity.

Here’s an example:

example_list = [1, 2, 3]
print(sum(sum(example_list[i:j+1]) for i in range(len(example_list)) for j in range(i, len(example_list))))

Output: 20

The one-liner uses a nested list comprehension to build and sum up all sublists in place, enabling a succinct expression of the brute force solution.

Summary/Discussion

  • Method 1: Brute Force Approach. Simple to implement but inefficient for large lists due to O(n^3) time complexity.
  • Method 2: Prefix Sum Technique. Offers a good balance between complexity and understanding, with a better O(n^2) time complexity.
  • Method 3: Cumulative Index Method. Greatly efficient O(n) solution utilizing the element’s positional contributions.
  • Method 4: Dynamic Programming. Makes use of previously computed sums; still O(n^2) but optimizes for computational repetition.
  • Bonus One-Liner Method 5: List Comprehension. Offers a concise solution but at the cost of computational efficiency similar to brute force.