LeetCode 1800: Maximum Ascending Subarray Sum
LeetCode 1800 Solution 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)
: :
LeetCode 1800 Solutions in Java, C++, Python
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);
}
}Interactive Code Editor for LeetCode 1800
Improve Your LeetCode 1800 Solution
Use the editor below to refine the provided solution for LeetCode 1800. Select a programming language and try the following:
- Add import statements if required.
- Optimize the code for better time or space complexity.
- Add test cases to validate edge cases and common scenarios.
- Handle error conditions or invalid inputs gracefully.
- Experiment with alternative approaches to deepen your understanding.
Click "Run Code" to execute your solution and view the output. If errors occur, check the line numbers and debug accordingly. Resize the editor by dragging its bottom edge.
Loading editor...