1800. Maximum Ascending Subarray Sum

Problem

Given an array of positive integers nums, return the **maximum possible sum of an *ascending* subarray in **nums.

A subarray is defined as a contiguous sequence of numbers in an array.

A subarray [numsl, numsl+1, ..., numsr-1, numsr] is ascending if for all i where l <= i < r, numsi < numsi+1. Note that a subarray of size 1 is ascending.

  Example 1:

Input: nums = [10,20,30,5,10,50]
Output: 65
Explanation: [5,10,50] is the ascending subarray with the maximum sum of 65.

Example 2:

Input: nums = [10,20,30,40,50]
Output: 150
Explanation: [10,20,30,40,50] is the ascending subarray with the maximum sum of 150.

Example 3:

Input: nums = [12,17,15,13,10,11,12]
Output: 33
Explanation: [10,11,12] is the ascending subarray with the maximum sum of 33.

  Constraints:

• 1 <= nums.length <= 100

• 1 <= nums[i] <= 100

Solution (Java)

class Solution {
    public int maxAscendingSum(int[] nums) {
        int maxSum = nums[0];
        int i = 0;
        int j = i + 1;
        while (i < nums.length - 1 && j < nums.length) {
            int sum = nums[j - 1];
            while (j < nums.length && nums[j] - nums[j - 1] > 0) {
                sum += nums[j];
                j++;
            }
            i = j;
            maxSum = Math.max(maxSum, sum);
            j++;
        }
        return maxSum;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).