How do you solve Maximum Ascending Subarray Sum on LeetCode?

Maximum Ascending Subarray Sum (LeetCode #1800) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The end of an ascending subarray marks the start of a new potential subarray.

What is the brute force solution for Maximum Ascending Subarray Sum?

The brute force approach for Maximum Ascending Subarray Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subarrays and calculating their sums. This is simple to implement but can be inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Ascending Subarray Sum?

Maximum Ascending Subarray Sum uses the following concepts: Array. The recommended patterns to study are: Two Pointers, Sliding Window, Array.

What are the interview tips for Maximum Ascending Subarray Sum?

Always consider edge cases, such as arrays with only one element. Explain your thought process clearly while coding. Practice dry-running your solution on paper before coding.

What are common mistakes when solving Maximum Ascending Subarray Sum?

Not resetting the current sum when the sequence is not ascending. Forgetting to update the maximum sum after checking the current sum.

#1800

Maximum Ascending Subarray Sum

Easy

ArrayTwo PointersSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the array, maintaining a running sum of the current ascending subarray and updating the maximum sum as needed. This is efficient and avoids unnecessary calculations.

⚙️

Algorithm

5 steps

1Step 1: Initialize variables for the maximum sum and the current sum.
2Step 2: Iterate through the array starting from the second element.
3Step 3: If the current element is greater than the previous one, add it to the current sum.
4Step 4: If not, reset the current sum to the current element.
5Step 5: Update the maximum sum if the current sum exceeds it.

solution.py9 lines

1def maxAscendingSum(nums):
2    max_sum = current_sum = nums[0]
3    for i in range(1, len(nums)):
4        if nums[i] > nums[i - 1]:
5            current_sum += nums[i]
6        else:
7            current_sum = nums[i]
8        max_sum = max(max_sum, current_sum)
9    return max_sum

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(1) since we only use a few variables for tracking sums.

1The end of an ascending subarray marks the start of a new potential subarray.
2Tracking the current sum allows for efficient updates without re-evaluating previous elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.