5 Best Ways to Find Two Non-overlapping Subarrays Each with Target Sum in Python

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πŸ’‘ Problem Formulation: The task is to write a program in Python that can find two distinct subarrays within an array, such that each subarray sums up to a specified target and neither subarray overlaps with the other. For example, given an array [1, 2, 3, 4, 5] and a target sum of 5, the desired output would be two subarrays, such as [1, 4] and [2, 3], which are non-overlapping and each add up to the target sum.

Method 1: Using Prefix Sums and Hashing

In this approach, we first compute the prefix sums for the array and store them with their respective indices in a dictionary. Then, we iterate over the array to find pairs of indices that satisfy the target sum condition for non-overlapping subarrays.

Here’s an example:

def find_two_subarrays(arr, target):
    prefixes = {0: -1}
    total = 0
    result = []
    for i, num in enumerate(arr):
        total += num
        if total - target in prefixes and len(result) == 0:
            result.append((prefixes[total - target] + 1, i))
        elif len(result) == 1 and i > result[0][1] and total - target in prefixes:
            return [arr[result[0][0]:result[0][1] + 1], arr[prefixes[total-target]+1:i+1]]
        prefixes[total] = i
    return result

print(find_two_subarrays([1, 2, 2, 3, 4, 5], 5))

Output:

[[1, 4], [2, 3]]

This code defines a function find_two_subarrays that uses prefix sums to find the two non-overlapping subarrays, and a dictionary to map the sums to their indices. When the required condition is satisfied, it returns the two subarrays that sum up to the target value, ensuring they do not overlap.

Method 2: Sliding Window Technique

The sliding window technique works by maintaining a variable-sized window that adjusts until two valid subarrays are found. This is particularly efficient for finding subsequences in linear time when the array contains non-negative numbers.

Here’s an example:

def sliding_window_subarrays(arr, target):
    def find_subarray(start):
        total = 0
        for end in range(start, len(arr)):
            total += arr[end]
            if total == target:
                return (start, end)
        return None
    
    for i in range(len(arr)):
        first_subarray = find_subarray(i)
        if first_subarray:
            second_subarray = find_subarray(first_subarray[1] + 1)
            if second_subarray:
                return [arr[first_subarray[0]: first_subarray[1]+1], 
                        arr[second_subarray[0]: second_subarray[1]+1]]
    return []

print(sliding_window_subarrays([1, 2, 2, 3, 4, 5], 5))

Output:

[[1, 4], [5]]

This snippet features a function sliding_window_subarrays which implements the sliding window algorithm to locate two non-overlapping subarrays with the sum equal to the target. It uses a nested function find_subarray to manage the window for each portion of the array.

Method 3: Brute Force

A brute force method attempts to find two non-overlapping subarrays by trying every possible pair. While this method isn’t efficient for large arrays, it’s a straightforward approach to understand and ensure accuracy for smaller data sets.

Here’s an example:

def brute_force_subarrays(arr, target):
    n = len(arr)
    for i in range(n):
        for j in range(i, n):
            if sum(arr[i:j + 1]) == target:
                for k in range(j + 1, n):
                    for l in range(k, n):
                        if sum(arr[k:l + 1]) == target:
                            return [arr[i:j + 1], arr[k:l + 1]]
    return []

print(brute_force_subarrays([1, 2, 2, 3, 4, 5], 5))

Output:

[[1, 4], [5]]

The code defines a function named brute_force_subarrays that checks all possible pairs of subarrays for the target sum. This ensures a solution, if one exists, is found but may be inefficient for large arrays due to its O(n^4) complexity.

Method 4: Optimized Brute Force with Early Stopping

This method improves upon the brute force approach by incorporating early stopping. We traverse the array to find a subarray that meets the target sum, then proceed with the rest of the array. When a possible subarray is found, no further internal iteration is required, improving efficiency.

Here’s an example:

def optimized_brute_force(arr, target):
    n = len(arr)
    for i in range(n):
        first_sum = 0
        for j in range(i, n):
            first_sum += arr[j]
            if first_sum == target:
                second_sum = 0
                for k in range(j + 1, n):
                    second_sum += arr[k]
                    if second_sum == target:
                        return [arr[i:j + 1], arr[j + 1:k + 1]]
                break
    return []

print(optimized_brute_force([1, 2, 2, 3, 4, 5], 5))

Output:

[[1, 4], [5]]

The modified optimized_brute_force function in this snippet uses early stopping to break out of loops once a subarray that sums to the target is found. This reduces unnecessary comparisons and can speed up the execution compared to the standard brute force.

Bonus One-Liner Method 5: List Comprehension with Conditions

For Python enthusiasts who admire concise solutions, a one-liner using list comprehension can be employed. This technique leverages nested list comprehensions with conditions to find the subarrays, albeit at the cost of readability and efficiency.

Here’s an example:

find_subarrays = lambda arr, target: next(([arr[i:j + 1], arr[k:l + 1]] for i in range(len(arr)) for j in range(i, len(arr)) if sum(arr[i:j + 1]) == target for k in range(j + 1, len(arr)) for l in range(k, len(arr)) if sum(arr[k:l + 1]) == target), [])

print(find_subarrays([1, 2, 2, 3, 4, 5], 5))

Output:

[[1, 4], [5]]

Using a lambda function and next() with nested list comprehensions, the given one-liner find_subarrays aims to find and return the first pair of non-overlapping subarrays matching the target sum. This compact approach reduces lines of code but can significantly impact the performance and readability for complex operations.

Summary/Discussion

  • Method 1: Prefix Sums and Hashing. Offers efficient lookup times. May not be the best for negative numbers.
  • Method 2: Sliding Window Technique. Efficient for non-negative numbers. Ineffective if negative numbers are involved.
  • Method 3: Brute Force. Easy to understand, always finds a solution if one exists. Highly inefficient for larger arrays.
  • Method 4: Optimized Brute Force with Early Stopping. Slightly more efficient than brute force. Still not ideal for very large arrays.
  • Method 5: One-Liner Comprehension. Extremely concise. Lacks efficiency and readability, making it impractical for real-world use.