> For the complete documentation index, see [llms.txt](https://liuxue2010.gitbook.io/data-structure-and-algorithms/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://liuxue2010.gitbook.io/data-structure-and-algorithms/array-and-numbers/maximum-subarray-ii.md).

# Maximum Subarray II

Given an array of integers, find two non-overlapping subarrays which have the largest sum.

The number in each subarray should be contiguous.

Return the largest sum.

## Example

For given \[1, 3, -1, 2, -1, 2], the two subarrays are \[1, 3] and \[2, -1, 2] or \[1, 3, -1, 2] and \[2], they both have the largest sum 7.

## Solution

两个范围内数组相加的最大值。

if the question is about two range then consider two pointers.

don't give total defualt value 0

don't do p1\[i] + p2\[i] due to the overlapping issue.

```
for i in range\(n - 1\):

    total = max\(total, p1\[i\] + p2\[i + 1\]\) no overlapping issue.

return total
```

```
class Solution:
    """
    @param nums: A list of integers
    @return: An integer denotes the sum of max two non-overlapping subarrays
    two pointers.
    """
    #两个范围内数组相加的最大值。
    def maxTwoSubArrays(self, nums):
        # write your code here    
        n = len(nums)

        result = sys.maxint * -1

        p1 = [0] * n
        p2 = [0] * n 

        total, sum = sys.maxint * -1, 0

        for i in range(n):
            sum += nums[i]
            total = max(total, sum)
            p1[i] = total
            sum = max(sum, 0)

        total, sum = sys.maxint * -1, 0
        for i in range(n - 1, -1, -1):
            sum += nums[i]
            total = max(total, sum)
            p2[i] = total
            sum = max(sum, 0)

        for i in range(n - 1):
            result = max(result, p1[i] + p2[i + 1])

        return result
```