// Source : https://leetcode.com/problems/maximum-ascending-subarray-sum/
// Author : Hao Chen
// Date   : 2021-03-21

/***************************************************************************************************** 
 *
 * 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.
 * 
 * Example 4:
 * 
 * Input: nums = [100,10,1]
 * Output: 100
 * 
 * Constraints:
 * 
 * 	1 <= nums.length <= 100
 * 	1 <= nums[i] <= 100
 ******************************************************************************************************/

class Solution {
public:
    int maxAscendingSum(vector<int>& nums) {
        int maxSum = nums[0];
        int sum = maxSum;
        for(int i=1; i<nums.size(); i++) {
            if (nums[i] > nums[i-1]) {
                sum += nums[i];
            }else{
                maxSum = maxSum < sum ? sum : maxSum;
                sum = nums[i];
            }
        }
        maxSum = maxSum < sum ? sum : maxSum;
        return maxSum;
    }
};