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 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.
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 and prefix sums. The idea is to precompute the sums of all subarrays of a given length and then use these precomputed sums to find the maximum sum of three non-overlapping subarrays efficiently.
Here is a step-by-step breakdown of the optimized algorithm:
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 up to each point
left = [0] * (n - k + 1)
right = [0] * (n - k + 1)
# Fill the left array
best = 0
for i in range(n - k + 1):
if sum_k[i] > sum_k[best]:
best = i
left[i] = best
# Fill the right array
best = n - k
for i in range(n - k, -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, n - 2 * k + 1):
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
The time complexity of this approach is O(n), where n is the length of the input array. This is because we make a constant number of passes through the array. The space complexity is also O(n) due to the additional arrays used for storing subarray sums and indices.
Potential edge cases include:
These edge cases are handled by the algorithm as it considers all possible subarrays of the given length.
To test the solution comprehensively, consider the following test cases:
# Test case 1: General case
assert maxSumOfThreeSubarrays([2, 3, -8, 7, -2, 9, -9, 7, -2, 4], 2) == 28
# Test case 2: All positive numbers
assert maxSumOfThreeSubarrays([1, 2, 3, 4, 5, 6, 7, 8, 9], 2) == 27
# Test case 3: All negative numbers
assert maxSumOfThreeSubarrays([-1, -2, -3, -4, -5, -6, -7, -8, -9], 2) == -12
# Test case 4: Mixed positive and negative numbers
assert maxSumOfThreeSubarrays([1, -2, 3, -4, 5, -6, 7, -8, 9], 2) == 15
# Test case 5: Optimal subarrays at the beginning
assert maxSumOfThreeSubarrays([9, 8, 7, 6, 5, 4, 3, 2, 1], 2) == 27
When approaching such problems, consider breaking them down into smaller subproblems and using dynamic programming to store intermediate results. Practice solving similar problems to develop a deeper understanding of the techniques involved.
Understanding and solving problems like the maximum sum of three non-overlapping subarrays is crucial for developing strong problem-solving skills. By breaking down the problem and using efficient algorithms, we can solve complex problems effectively.
For further reading and practice, consider the following resources: