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.

https://leetcode.com/problems/maximum-ascending-subarray-sum/

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

Test Cases

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class Solution:
    def maxAscendingSum(self, nums: List[int]) -> int:

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import pytest

from solution import Solution


@pytest.mark.parametrize('nums, expected', [
    ([10,20,30,5,10,50], 65),
    ([10,20,30,40,50], 150),
    ([12,17,15,13,10,11,12], 33),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected):
    assert sol.maxAscendingSum(nums) == expected

Thoughts

直接遍历数组，记录当前连续递增子数组之和，如果下一个数字仍然是递增则累加到和，否则重置和。

时间复杂度 O(n)，空间复杂度 O(1)。

Code

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class Solution:
    def maxAscendingSum(self, nums: list[int]) -> int:
        max2 = lambda a, b: a if a >= b else b
        max_sum = 0
        s = 0
        prev = nums[0]
        for num in nums:
            if num > prev:
                s += num
            else:
                s = num

            max_sum = max2(max_sum, s)
            prev = num

        return max_sum

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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