LeetCode in Kotlin
1800. Maximum Ascending Subarray Sum
Easy
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
Solution
class Solution {
fun maxAscendingSum(nums: IntArray): Int {
var maxSum = nums[0]
var i = 0
var j = i + 1
while (i < nums.size - 1 && j < nums.size) {
var sum = nums[j - 1]
while (j < nums.size && nums[j] - nums[j - 1] > 0) {
sum += nums[j]
j++
}
i = j
maxSum = Math.max(maxSum, sum)
j++
}
return maxSum
}
}