Maximum Ascending Subarray Sum - Video Solutions

Problem Overview: You are given an integer array and need to compute the maximum possible sum of a contiguous subarray where each element is strictly greater than the previous one. The subarray must remain strictly ascending from left to right, and the goal is to track the highest sum among all such segments.

This problem falls under common array traversal patterns. The key challenge is recognizing when an ascending streak continues and when it breaks so you can reset the running total.

Approach 1: Single Pass with Current Sum Tracker (O(n) time, O(1) space)

Scan the array once while maintaining two variables: currentSum for the sum of the current ascending segment and maxSum for the best result seen so far. Start with the first element as the current sum. While iterating, check if nums[i] > nums[i-1]. If true, the ascending order continues, so add the element to currentSum. If not, the ascending streak breaks and you reset currentSum to the current element. After each step, update maxSum = max(maxSum, currentSum). This works because every ascending segment is processed exactly once, and each element contributes to at most one running sum.

The insight is that an ascending subarray cannot extend across a non-increasing boundary. Instead of checking every possible subarray, you treat each increasing run as a single candidate segment. The algorithm performs a single linear pass through the array, giving O(n) time complexity and constant O(1) space.

Approach 2: Divide and Conquer Strategy (O(n log n) time, O(log n) space)

This approach recursively splits the array into two halves, similar to classic divide-and-conquer patterns. For each segment, compute three values: the best ascending subarray entirely in the left half, the best entirely in the right half, and a potential ascending subarray that crosses the midpoint. The crossing case is valid only if the boundary elements maintain the strictly increasing condition.

During the merge step, you combine results from both halves while checking whether the rightmost value of the left segment is smaller than the leftmost value of the right segment. If so, the ascending sequence can extend across the midpoint, and you compute the combined sum. Recursion depth contributes O(log n) stack space, and each level processes O(n) work, leading to O(n log n) total time.

Recommended for interviews: The single-pass running sum approach is what interviewers expect. It demonstrates strong understanding of array traversal and state tracking while achieving optimal O(n) time and O(1) space. The divide-and-conquer version is academically interesting and shows familiarity with recursive decomposition, but it introduces unnecessary overhead for this problem.