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:
The core challenge of this problem is to find three non-overlapping subarrays that together yield the maximum possible sum. This problem is significant in scenarios where we need to maximize profit or minimize cost by selecting optimal segments from a sequence of data points.
Potential pitfalls include overlapping subarrays and not considering all possible subarray combinations.
To solve this problem, we can break it down into manageable steps:
A naive solution would involve checking all possible combinations of three subarrays, but this approach is not optimal due to its high time complexity.
We can optimize the solution using dynamic programming:
Here is a step-by-step breakdown of the optimized algorithm:
def max_sum_of_three_subarrays(nums):
n = len(nums)
k = 3 # Length of each subarray
prefix_sums = [0] * (n + 1)
# Calculate prefix sums
for i in range(n):
prefix_sums[i + 1] = prefix_sums[i] + nums[i]
# Initialize arrays to store the best sums
left = [0] * n
right = [0] * n
max_sum = 0
# Calculate the best sum for the left subarray
for i in range(k - 1, n):
current_sum = prefix_sums[i + 1] - prefix_sums[i + 1 - k]
if i == k - 1 or current_sum > prefix_sums[left[i - 1] + k] - prefix_sums[left[i - 1]]:
left[i] = i + 1 - k
else:
left[i] = left[i - 1]
# Calculate the best sum for the right subarray
for i in range(n - k, -1, -1):
current_sum = prefix_sums[i + k] - prefix_sums[i]
if i == n - k or current_sum >= prefix_sums[right[i + 1] + k] - prefix_sums[right[i + 1]]:
right[i] = i
else:
right[i] = right[i + 1]
# Calculate the maximum sum of three non-overlapping subarrays
for i in range(k, n - 2 * k + 1):
l = left[i - 1]
r = right[i + k]
current_sum = (prefix_sums[l + k] - prefix_sums[l] +
prefix_sums[i + k] - prefix_sums[i] +
prefix_sums[r + k] - prefix_sums[r])
max_sum = max(max_sum, current_sum)
return max_sum
# Example usage
nums = [2, 3, -8, 7, -2, 9, -9, 7, -2, 4]
print(max_sum_of_three_subarrays(nums)) # Output: 28
The time complexity of this solution is O(n), where n is the length of the input array. This is because we iterate through the array a constant number of times. The space complexity is also O(n) due to the prefix sums and the arrays used to store the best sums.
Potential edge cases include:
These edge cases are handled by the algorithm as it considers all possible subarray positions.
To test the solution comprehensively, consider the following test cases:
Use testing frameworks like unittest
or pytest
to automate the testing process.
When approaching such problems, consider breaking them down into smaller subproblems. Use dynamic programming to store intermediate results and avoid redundant calculations. Practice similar problems to improve your problem-solving skills.
Understanding and solving problems like this one is crucial for developing strong algorithmic thinking. Practice regularly and explore different approaches to enhance your skills.