# LeetCode 1800. 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`

## Solution

```
class Solution {
public:
    int maxAscendingSum(vector<int>& nums) {
        int ans = nums[0];
        
        int cur = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] > nums[i-1]) {
                cur += nums[i];
            } else {
                cur = nums[i];
            }
            ans = max(ans, cur);
        }
        
        return ans;
    }
};
```
