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 smaller steps:
Let's start with a naive solution and then optimize it.
The naive solution involves checking all possible combinations of three subarrays, which is computationally expensive and not feasible for large arrays.
We can optimize the solution using dynamic programming:
Here is a step-by-step breakdown of the optimized algorithm:
def maxSumOfThreeSubarrays(nums, k):
# Step 1: Calculate the prefix sums
n = len(nums)
prefix_sums = [0] * (n + 1)
for i in range(n):
prefix_sums[i + 1] = prefix_sums[i] + nums[i]
# Step 2: Initialize arrays to store the maximum sums
left = [0] * n
right = [0] * n
max_sum = 0
# Step 3: Calculate the maximum sum of subarrays ending at or before each index
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]
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]
# Step 4: Find the maximum sum of three non-overlapping subarrays
for i in range(k, n - 2 * k + 1):
l, r = left[i - 1], 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]
k = 2
print(maxSumOfThreeSubarrays(nums, k)) # Output: 28
The time complexity of this solution 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 prefix sums and tracking maximum subarray sums.
Potential edge cases include:
Each of these cases should be tested to ensure the algorithm handles them correctly.
To test the solution comprehensively, consider the following test cases:
# Test case 1: Mixed positive and negative numbers nums = [2, 3, -8, 7, -2, 9, -9, 7, -2, 4] k = 2 assert maxSumOfThreeSubarrays(nums, k) == 28 # Test case 2: All positive numbers nums = [1, 2, 3, 4, 5, 6, 7, 8, 9] k = 3 assert maxSumOfThreeSubarrays(nums, k) == 45 # Test case 3: All negative numbers nums = [-1, -2, -3, -4, -5, -6, -7, -8, -9] k = 2 assert maxSumOfThreeSubarrays(nums, k) == -12 # Test case 4: Minimum length array nums = [1, 2, 3] k = 1 assert maxSumOfThreeSubarrays(nums, k) == 6
When approaching such problems, consider breaking them down into smaller, manageable steps. 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. By breaking down the problem and using dynamic programming, we can find an efficient solution. Practice and exploration of similar problems will further enhance your skills.