Approach: Reset on non-ascending

def maxAscendingSum(nums):
    best = cur = nums[0]
    for i in range(1, len(nums)):
        cur = cur + nums[i] if nums[i] > nums[i-1] else nums[i]
        best = max(best, cur)
    return best

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def maxAscendingSum(self, nums):
        best = cur = nums[0]
        for i in range(1, len(nums)):
            cur = cur + nums[i] if nums[i] > nums[i-1] else nums[i]
            best = max(best, cur)
        return best

import java.util.*;
class Solution {
    public int maxAscendingSum(int[] nums) {
        int best = nums[0], cur = nums[0];
        for (int i = 1; i < nums.length; i++) {
            cur = nums[i] > nums[i-1] ? cur + nums[i] : nums[i];
            best = Math.max(best, cur);
        }
        return best;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int maxAscendingSum(vector<int>& nums) {
        int best = nums[0], cur = nums[0];
        for (int i = 1; i < (int)nums.size(); i++) {
            cur = nums[i] > nums[i-1] ? cur + nums[i] : nums[i];
            best = max(best, cur);
        }
        return best;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var maxAscendingSum = function(nums) {
    let best = nums[0], cur = nums[0];
    for (let i = 1; i < nums.length; i++) {
        cur = nums[i] > nums[i-1] ? cur + nums[i] : nums[i];
        if (cur > best) best = cur;
    }
    return best;
};

Maximum Ascending Subarray Sum

Example 1

Example 2

Example 3

Constraints