Solution to the problem of the longest subarray sum... That is, the two ascending... Executable version: https://onlinegdb.com/jYaDXgI5T More information at https://leetcode.com/problems/maximum-ascending-subarray-sum/

LargestAscendingSubArraySum.java

/**
 *  @author Marcello.DeSales@gmail.com 
 * 
 *  Solution to the problem of the longest subarray sum... That is, the two ascending arrays
 * [10,20,30,5,10,50] = 65     [5,10,50]
 * [12,17,15,13,10,11,12] = 33.       [10,11,12]
 **/
public class Main {

   public static void main(String[] args) {
        int[] nums = new int[]{12,17,15,13,10,11,12};

        int maxComputed = computeSum(nums);
        System.out.println("value: " + maxComputed);

        if (maxComputed == 65) {
            System.out.println("Success");
        }
    }

    private static int computeSum(int[] nums) {
        if (nums.length == 1) return nums[0];
        int currentSum = 0;
        leftWindow: for (int left = 0; left < nums.length - 1; left++) {
            int leftValue = nums[left];

            // the local sum is already the left side
            int localSum = leftValue;
            for (int right = left + 1; right < nums.length; right++) {
                int rightValue = nums[right];
                if (leftValue < rightValue) {
                    // sum is already the left side
                    localSum += rightValue;

                } else if (right == nums.length - 1) {
                    // When we hit the right side of the window, we know we don't have subarrays
                    currentSum = currentSum >= localSum ? currentSum : localSum;
                    break leftWindow;

                } else {
                    // When the value is smaller, we know we don't have an ascending sequence anymore
                    // The left side turns to be the next value of the right side
                    left = right - 1;
                    break;
                }
                leftValue = nums[right];
            }
            // As the current local subarray sum now is found, the currentSum can be compared with the previous
            currentSum = currentSum >= localSum ? currentSum : localSum;
            localSum = 0;
        }

        return currentSum;
    }
}

run.log

[10,20,30,5,10,50]
value: 65
Success

Process finished with exit code 0

[12,17,15,13,10,11,12]
value: 33
Success

Process finished with exit code 0