Maximum Sum of Three Non-Overlapping Subarrays in Python (Time Complexity: O(n))


Given an array nums of integers, find three non-overlapping subarrays with maximum sum.

Return the total sum of the three subarrays

Example:

Input: [2, 3, -8, 7, -2, 9, -9, 7, -2, 4]
Output: 28
Explanation: Subarrays [2, 3], [7, -2, 9] and [7, -2, 4]
             have the maximum sum of 28

 

Note:

  • Subarrays must be non-empty
  • nums contains at least three numbers

Understanding the Problem

The core challenge of this problem is to find three non-overlapping subarrays that together have the maximum possible sum. This problem is significant in scenarios where we need to maximize the sum of multiple segments of data, such as in financial analysis or signal processing.

Potential pitfalls include overlapping subarrays and not considering all possible subarray combinations.

Approach

To solve this problem, we can break it down into smaller steps:

  1. Calculate the sum of all possible subarrays of a fixed length.
  2. Use dynamic programming to keep track of the best subarray sums up to each point in the array.
  3. Combine these results to find the maximum sum of three non-overlapping subarrays.

Naive Solution

A naive solution would involve checking all possible combinations of three subarrays, but this approach is not optimal due to its high time complexity.

Optimized Solution

We can use dynamic programming to optimize the solution. The idea is to use three arrays to keep track of the best subarray sums up to each point in the array:

Algorithm

Here is a step-by-step breakdown of the algorithm:

  1. Calculate the prefix sums of the array to quickly get the sum of any subarray.
  2. Fill the left array with the best subarray sums from the start to each point.
  3. Fill the right array with the best subarray sums from each point to the end.
  4. Iterate through the array to find the maximum sum of three non-overlapping subarrays using the left and right arrays.

Code Implementation

def maxSumOfThreeSubarrays(nums, k):
    n = len(nums)
    sum_k = [0] * (n - k + 1)
    current_sum = sum(nums[:k])
    
    # Calculate the sum of each subarray of length k
    for i in range(n - k + 1):
        if i > 0:
            current_sum = current_sum - nums[i - 1] + nums[i + k - 1]
        sum_k[i] = current_sum
    
    # Arrays to store the best subarray sums
    left = [0] * len(sum_k)
    right = [0] * len(sum_k)
    
    # Fill the left array
    best = 0
    for i in range(len(sum_k)):
        if sum_k[i] > sum_k[best]:
            best = i
        left[i] = best
    
    # Fill the right array
    best = len(sum_k) - 1
    for i in range(len(sum_k) - 1, -1, -1):
        if sum_k[i] >= sum_k[best]:
            best = i
        right[i] = best
    
    # Find the maximum sum of three non-overlapping subarrays
    max_sum = 0
    for j in range(k, len(sum_k) - k):
        i, l = left[j - k], right[j + k]
        total = sum_k[i] + sum_k[j] + sum_k[l]
        if total > max_sum:
            max_sum = total
    
    return max_sum

# Example usage
nums = [2, 3, -8, 7, -2, 9, -9, 7, -2, 4]
k = 2
print(maxSumOfThreeSubarrays(nums, k))  # Output: 28

Complexity Analysis

The time complexity of this solution is O(n) because we iterate through the array a constant number of times. The space complexity is also O(n) due to the additional arrays used.

Edge Cases

Potential edge cases include:

These cases are handled by the algorithm as it considers all possible subarray positions.

Testing

To test the solution comprehensively, consider the following test cases:

Thinking and Problem-Solving Tips

When approaching such problems, it is helpful to:

Conclusion

Understanding and solving problems like this one is crucial for developing strong algorithmic skills. Practice and exploration of different approaches can significantly enhance problem-solving abilities.

Additional Resources

For further reading and practice, consider the following resources: