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1800. Maximum Ascending Subarray Sum

Description

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

Solutions

Solution 1: Direct Simulation

We use a variable $t$ to record the current sum of the ascending subarray, and a variable $ans$ to record the maximum sum of the ascending subarray.

Traverse the array $nums$:

If the current element is the first element of the array, or the current element is greater than the previous one, then add the current element to the sum of the current ascending subarray, i.e., $t = t + nums[i]$, and update the maximum sum of the ascending subarray $ans = \max(ans, t)$. Otherwise, the current element does not satisfy the condition of the ascending subarray, so reset the sum $t$ of the current ascending subarray to the current element, i.e., $t = nums[i]$.

After the traversal, return the maximum sum of the ascending subarray $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

  • class Solution {
    public int maxAscendingSum(int[] nums) {
        int ans = 0, t = 0;
        for (int i = 0; i < nums.length; ++i) {
            if (i == 0 || nums[i] > nums[i - 1]) {
                t += nums[i];
                ans = Math.max(ans, t);
            } else {
                t = nums[i];
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int maxAscendingSum(vector<int>& nums) {
        int ans = 0, t = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (i == 0 || nums[i] > nums[i - 1]) {
                t += nums[i];
                ans = max(ans, t);
            } else {
                t = nums[i];
            }
        }
        return ans;
    }
};

  • class Solution:
    def maxAscendingSum(self, nums: List[int]) -> int:
        ans = t = 0
        for i, v in enumerate(nums):
            if i == 0 or v > nums[i - 1]:
                t += v
                ans = max(ans, t)
            else:
                t = v
        return ans


  • func maxAscendingSum(nums []int) int {
	ans, t := 0, 0
	for i, v := range nums {
		if i == 0 || v > nums[i-1] {
			t += v
			if ans < t {
				ans = t
			}
		} else {
			t = v
		}
	}
	return ans
}

  • function maxAscendingSum(nums: number[]): number {
    const n = nums.length;
    let res = nums[0];
    let sum = nums[0];
    for (let i = 1; i < n; i++) {
        if (nums[i] <= nums[i - 1]) {
            res = Math.max(res, sum);
            sum = 0;
        }
        sum += nums[i];
    }
    return Math.max(res, sum);
}


  • impl Solution {
    pub fn max_ascending_sum(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut res = nums[0];
        let mut sum = nums[0];
        for i in 1..n {
            if nums[i - 1] >= nums[i] {
                res = res.max(sum);
                sum = 0;
            }
            sum += nums[i];
        }
        res.max(sum)
    }
}

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