# LeetCode 1800. Maximum Ascending Subarray Sum

## Description

https://leetcode.com/problems/maximum-ascending-subarray-sum/

Given an array of positive integers `nums`, return the maximum possible sum of an ascending subarray in `nums`.

A subarray is defined as a contiguous sequence of numbers in an array.

A subarray `[numsl, numsl+1, ..., numsr-1, numsr]` is ascending if for all `i` where `l <= i < r``numsi < numsi+1`. Note that a subarray of size `1` is ascending.

Example 1:

```Input: nums = [10,20,30,5,10,50]
Output: 65
Explanation: [5,10,50] is the ascending subarray with the maximum sum of 65.
```

Example 2:

```Input: nums = [10,20,30,40,50]
Output: 150
Explanation: [10,20,30,40,50] is the ascending subarray with the maximum sum of 150.
```

Example 3:

```Input: nums = [12,17,15,13,10,11,12]
Output: 33
Explanation: [10,11,12] is the ascending subarray with the maximum sum of 33.
```

Example 4:

```Input: nums = [100,10,1]
Output: 100
```

Constraints:

• `1 <= nums.length <= 100`
• `1 <= nums[i] <= 100`

## Explanation

Keep track of sub array num and update the maximum sub array sum if the sub array sum is greater than maximum sub array sum.

## Python Solution

``````class Solution:
def maxAscendingSum(self, nums: List[int]) -> int:
max_sum = 0

i = 0
prev = None

sub_sum = 0
while i < len(nums):
num = nums[i]

if not prev or num > prev:
sub_sum += num
max_sum = max(max_sum, sub_sum)
else:
sub_sum  = num

prev = num
i += 1

return max_sum

``````
• Time Complexity: O(N).
• Space Complexity: O(1).