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 <= 1001 <= 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:
\( O(1) \).