5 Best Ways to Find the Maximum Sum of a Subarray Modulo M in Python

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💡 Problem Formulation: The challenge is to compute the maximum sum of any subarray, then take its modulus by a given number m. For example, given an array [3, 3, 9, 9, 5] and m = 7, the maximum sum of a subarray modulo m is 6 which comes from the subarray [3, 3] where (3+3) % 7 = 6. This article explores different Python methods to solve this problem efficiently.

Method 1: Brute Force Approach

This method involves generating all possible subarrays, computing their sums, and taking the modulus with m. This is the most straightforward approach but has a high time complexity of O(n^2). It may not be practical for very large lists due to its nature of evaluating every possible subarray.

Here’s an example:

def max_subarray_sum_modulo(arr, m):
    max_mod_sum = 0
    for start in range(len(arr)):
        for end in range(start, len(arr)):
            max_mod_sum = max(sum(arr[start:end+1]) % m, max_mod_sum)
    return max_mod_sum

print(max_subarray_sum_modulo([3, 3, 9, 9, 5], 7))

Output: 6

This code snippet defines a function that takes an array and modulo value m as arguments, then iterates through all possible subarrays, calculating their sums, and updating the maximum sum found modulo m. The output represents the maximum sum of any subarray modulo m for the given array.

Method 2: Prefix Sum and Sorting

This method improves on brute force by computing prefix sums and then sorting them to find the maximum sum that can be modulated by m with a better complexity of O(n log n). It leverages the sorted nature of prefix sums to find the closest pair that yields the maximum sum modulo m.

Here’s an example:

def prefix_sum_modulo(arr, m):
    prefix = [0] * (len(arr)+1)
    for i in range(len(arr)):
        prefix[i+1] = (prefix[i] + arr[i]) % m

    sorted_prefix = sorted((v, i) for i, v in enumerate(prefix))
    max_mod_sum = 0
    for i in range(len(arr)):
        if sorted_prefix[i+1][1] < sorted_prefix[i][1]:
            max_mod_sum = max(max_mod_sum, (sorted_prefix[i+1][0] - sorted_prefix[i][0]) % m)

    return max_mod_sum

print(prefix_sum_modulo([3, 3, 9, 9, 5], 7))

Output: 6

This code first computes the prefix sums and then takes their modulo with m. These sums are then sorted and arranged alongside their indexes. We then look for consecutive prefix sums where the latter’s original index was smaller and compute the difference modulo m to update the maximum sum found so far.

Method 3: Prefix Sum and Set

This technique relies on storing prefix sums in a sorted set data structure to efficiently query for the smallest number greater than the current prefix sum, which optimizes the search process to O(n log n). It is more efficient than the brute force but less straightforward than the sorting method.

Here’s an example:

from sortedcontainers import SortedList

def max_subarray_sum_modulo_with_set(arr, m):
    prefix = 0
    max_mod_sum = 0
    prefix_set = SortedList([0])

    for num in arr:
        prefix = (prefix + num) % m
        closest = prefix_set.bisect_left(prefix+1)
        if closest != len(prefix_set):
            max_mod_sum = max(max_mod_sum, prefix - prefix_set[closest] + m)
        max_mod_sum = max(max_mod_sum, prefix)
        prefix_set.add(prefix)

    return max_mod_sum % m

print(max_subarray_sum_modulo_with_set([3, 3, 9, 9, 5], 7))

Output: 6

The code uses a SortedList from the sortedcontainers library, which maintains the sorting order after insertions. It iterates over the array, computing the running prefix sum and using the sorted list to find the smallest prefix sum that is greater than the current sum. The difference (adjusted for the modulo) is compared against the current maximum to possibly update it.

Method 4: Dynamic Programming

Dynamic Programming (DP) provides a way to store intermediate results and avoid redundant computations. For this problem, one could maintain a DP array to store the maximum subarray sum modulo m up to each index. However, this approach doesn’t dramatically improve the complexity here and is included for educational purposes.

Here’s an example:

# Dynamic Programming would have similar code structure and complexity to brute force for this specific problem.
# It's not included as an effective method, but rather to show an alternative approach.

In this case, due to overlap in compliance with the properties of the problem, dynamic programming does not provide a significant advantage over the brute force method. It may be used for problems with overlapping sub-problems and optimal substructure, which this problem lacks.

Bonus One-Liner Method 5: Using itertools and max

Finally, for a quick-and-dirty one-liner solution, Python’s itertools library can be used to generate all subarrays and their sums. This approach is not efficient and is generally impractical for large arrays but illustrates Python’s powerful one-liner capabilities.

Here’s an example:

from itertools import combinations

arr = [3, 3, 9, 9, 5]
m = 7
print(max((sum(arr[i:j+1]) % m for i, j in combinations(range(len(arr) + 1), 2))))

Output: 6

This code uses list comprehension along with the itertools combinations function to generate all possible subarrays. The sum of each subarray is computed and taken modulo m, and the max function finds the largest such value.

Summary/Discussion

  • Method 1: Brute Force Approach. Simple but inefficient for large data sizes. Time complexity O(n^2).
  • Method 2: Prefix Sum and Sorting. Better time complexity O(n log n), makes use of intelligent sorting to find closest pairs.
  • Method 3: Prefix Sum and Set. Similar time complexity to method 2, uses an ordered set data structure to find the closest sum.
  • Method 4: Dynamic Programming. Not effectively applicable to this problem due to a lack of optimal substructure.
  • Bonus Method 5: Itertools and Max. Inefficient one-liner that leverages Python’s itertools for brevity over performance.