1927. Maximum Ascending Subarray Sum¶

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1927. 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 <= 100
• 1 <= nums[i] <= 100

Solution¶

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

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here