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] <= 100class Solution {
public int maxAscendingSum(int[] nums) {
int n = nums.length, maxSubArraySum = nums[0], currSubArraySum = nums[0];
for (int i = 1; i < n; i++) {
if (nums[i - 1] < nums[i])
currSubArraySum += nums[i];
else {
maxSubArraySum = Math.max(maxSubArraySum, currSubArraySum);
currSubArraySum = nums[i];
}
}
return Math.max(maxSubArraySum, currSubArraySum);
}
}