1800

1800. Maximum Ascending Subarray Sum (Easy)

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

Related Topics:
Two Pointers

Solution 1.

// OJ: https://leetcode.com/problems/maximum-ascending-subarray-sum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxAscendingSum(vector<int>& A) {
        int ans = A[0], sum = A[0];
        for (int i = 1; i < A.size(); ++i) {
            if (A[i] > A[i - 1]) {
                sum += A[i];
            } else {
                sum = A[i];
            }
            ans = max(ans, sum);
        }
        return ans;
    }
};