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

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxAscendingSum = function(nums) {
    var maxSum = nums[0];
    var currSum = nums[0];
    for (var i = 1; i < nums.length; i++) {
        if (nums[i - 1] < nums[i]) {
            currSum += nums[i];
        } else {
            maxSum = Math.max(maxSum, currSum);
            currSum = nums[i];
        }
    }
    return Math.max(maxSum, currSum);
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).