LeetCode 1800: Maximum Ascending Subarray Sum Solution

Master LeetCode problem 1800 (Maximum Ascending Subarray Sum), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1800. Maximum Ascending Subarray Sum

Easy

Array

Problem Explanation

Explanation:

To solve this problem, we can iterate through the given array nums and keep track of the current sum of the ascending subarray. If the current element is greater than the previous element, we continue to add it to the sum. If not, we update the maximum sum so far if the current sum is greater. We then reset the current sum to the current element.

At each step, we compare the current sum with the maximum sum seen so far and update the maximum sum if needed. Finally, we return the maximum sum as the result.

Time Complexity: O(n) where n is the number of elements in the input array nums. Space Complexity: O(1)

: :

Solution Code

class Solution {
    public int maxAscendingSum(int[] nums) {
        int maxSum = nums[0];
        int currentSum = nums[0];

        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > nums[i - 1]) {
                currentSum += nums[i];
            } else {
                maxSum = Math.max(maxSum, currentSum);
                currentSum = nums[i];
            }
        }

        return Math.max(maxSum, currentSum);
    }
}

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