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
Approach 01:
-
C++
-
Python
#include <vector> #include <algorithm> using namespace std; class Solution { public: int maxAscendingSum(vector<int>& nums) { int currentSum = nums[0]; int maxSum = INT_MIN; int arraySize = nums.size(); for (int i = 1; i < arraySize; i++) { if (nums[i] > nums[i - 1]) { currentSum += nums[i]; } else { maxSum = max(maxSum, currentSum); currentSum = nums[i]; } } return max(maxSum, currentSum); } };
from typing import List class Solution: def maxAscendingSum(self, nums: List[int]) -> int: currentSum = nums[0] maxSum = float('-inf') arraySize = len(nums) for i in range(1, arraySize): if nums[i] > nums[i - 1]: currentSum += nums[i] else: maxSum = max(maxSum, currentSum) currentSum = nums[i] return max(maxSum, currentSum)
Time Complexity:
- Single Pass:
We iterate through the array once, processing each element in \( O(1) \) time.
- Overall Time Complexity:
\( O(N) \), where \( N \) is the size of the input array.
Space Complexity:
- Auxiliary Space:
We use only a few integer variables (
currentSum
,maxSum
,arraySize
), all of which take \( O(1) \) space. - Overall Space Complexity:
\( O(1) \).