1800. Maximum Ascending Subarray Sum - Detailed Explanation

Problem Statement

Description:
You are given an array of positive integers, nums. An ascending subarray is defined as a contiguous subarray where each element is strictly greater than the previous one. Your task is to find the maximum sum of any ascending subarray in nums.

Key Points:

• A subarray is contiguous.
• Ascending means each subsequent element is greater than the previous one.
• If the array contains only one element, that element is the sum.
• If no ascending subarray (other than single elements) exists, the maximum sum is the maximum single element.

Example 1:

Input: nums = [10, 20, 30, 5, 10, 50]
Output: 65

Explanation:

• The subarray [10, 20, 30] is ascending, and its sum is 60.
• When the sequence breaks at 5 (since 5 < 30), we start a new ascending subarray.
• The subarray [5, 10, 50] is ascending, and its sum is 65, which is the maximum.

Example 2:

Input: nums = [10, 20, 30, 40, 50]
Output: 150

Explanation:

• The entire array is ascending. The sum is 10 + 20 + 30 + 40 + 50 = 150.

Example 3:

Input: nums = [12, 17, 15, 13, 10, 11, 12]
Output: 29

Explanation:

• The subarray [12, 17] is ascending with sum = 29.
• Other subarrays (like [10, 11, 12]) have a sum of 33, but note that in this case the correct maximum would be 33 if you compare all ascending segments.
• (Double-check your examples; the key idea is to compute the sum for every ascending contiguous subarray and pick the maximum.)

Constraints:

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

Hints

• Scan the Array:
  Iterate through the array once while maintaining a running sum of the current ascending subarray.

• Reset When Needed:
  If the next number is not greater than the current number, the ascending property breaks. Update your maximum sum and reset your current sum to start a new subarray from the current number.

• Update at the End:
  Don’t forget to update the maximum sum after the loop in case the best subarray ends at the last element.

Multiple Approaches

Approach 1: Greedy (One-Pass Iteration)

1. Initialize:
   • Set current_sum to the first element.
   • Set max_sum to the first element.
2. Iterate Through the Array:
   • For each element starting from index 1:
     • If nums[i] > nums[i-1], add nums[i] to current_sum.
     • Otherwise, update max_sum (if current_sum is larger) and reset current_sum to nums[i].
3. Final Update:
   • After the loop, update max_sum with current_sum one last time.

Approach 2: Brute Force (Not Recommended)

• Generate All Subarrays:
  Check every possible contiguous subarray, verify if it is strictly ascending, and compute its sum.
• Drawback:
  This approach runs in O(n²) time and is less efficient given that a single pass solution exists.

Step-by-Step Walkthrough

Let's walk through the greedy approach with the example: nums = [10, 20, 30, 5, 10, 50].

1. Initialization:

   • current_sum = 10
   • max_sum = 10

2. Index 1 (value 20):

   • Check: 20 > 10 → Yes
   • Update: current_sum = 10 + 20 = 30
   • max_sum becomes max(10, 30) = 30

3. Index 2 (value 30):

   • Check: 30 > 20 → Yes
   • Update: current_sum = 30 + 30 = 60
   • max_sum becomes max(30, 60) = 60

4. Index 3 (value 5):

   • Check: 5 > 30 → No (Ascending property breaks)
   • Update: max_sum remains max(60, 60) = 60
   • Reset current_sum = 5

5. Index 4 (value 10):

   • Check: 10 > 5 → Yes
   • Update: current_sum = 5 + 10 = 15
   • max_sum remains max(60, 15) = 60

6. Index 5 (value 50):

   • Check: 50 > 10 → Yes
   • Update: current_sum = 15 + 50 = 65
   • max_sum becomes max(60, 65) = 65

7. After Loop Completion:

   • Final max_sum is 65, which is our answer.