1800: Maximum-Ascending-Subarray-Sum
Easy

table of contents

Just iterate through nums, keeping track of the current sum using counter when nums[i-1] < nums[i] and the maximum sum using ans by making it equal to the largest value between ans and counter everytime the current subarray stops ascending.

code

class Solution {
public:
    int longestMonotonicSubarray(vector<int>& nums) {
        int ans = 1;
        int counter = 1;
        int direction = 0;
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i-1] != nums[i]) {
                if (direction == ((nums[i-1] < nums[i]) ? 1 : -1)) {
                    ++counter;
                } else {
                    direction = (nums[i-1] < nums[i]) ? 1 : -1;
                    ans = max(ans, counter);
                    counter = 2;
                }
            } else {
                direction = 0;
                ans = max(ans, counter);
                counter = 0;
            }
        }
        ans = max(ans, counter);
        return ans;
    }
};

complexity