1800. Maximum Ascending Subarray Sum

Problem Description

The task is to find the highest sum of a contiguous subarray within an array of positive integers nums. Each subarray must be sorted in an ascending order without breaks, which means each element in the subarray must be strictly greater than its preceding element. Note that a subarray composed of a single element counts as ascending.

Intuition

The intuition behind the solution is to iterate through the array and maintain a running total t whenever the ascending condition is met. As soon as we hit a number that is not greater than the previous number, we must begin a new subarray and reset our running total to the current number's value because an ascending subarray can no longer continue past this point.

While iterating, we use the variable ans to keep track of the maximum sum observed so far. If our current running total t surpasses ans, we update ans with the value of t. By using a single pass through the array and updating these two variables appropriately, we can find the maximum sum of an ascending subarray efficiently.

Solution Approach

The solution is implemented in Python and follows a simple iterative approach to solve the problem. The code utilizes no extra data structures, operating directly on the input list to keep track of two important values:

t: This variable holds the current sum of the latest ascending subarray being considered.
ans: This is used to maintain the maximum sum encountered so far as we iterate through the array.

The iteration starts from the beginning of the list nums. For each number v at index i, we check whether it maintains an ascending order with respect to the previous number nums[i - 1]. The initial condition i == 0 accounts for the start of the array, and by default, we begin with an ascending subarray consisting of the first element.

If the current number v is greater than the previous number, we are still in an ascending subarray. We add v to our current sum t and then compare it with ans to potentially update the maximum sum found. This comparison and potential update take place using the max() function:

ans = max(ans, t)

When the ascending condition breaks (the current number is not greater than the previous one), we need to reset the running sum t to the current number's value, as we are now starting a new ascending subarray:

t = v

This process continues until we've examined each number in the array. By the end of the iteration, ans contains the maximum sum of an ascending subarray, which is then returned.

The overall algorithm exhibits O(n) time complexity, where n is the number of elements in input array, as it requires a single pass through the list. No additional space is used beyond the input and constant variables, which results in an O(1) space complexity.

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Example Walkthrough

Let's illustrate the solution approach with a small example. Consider the following array of positive integers:

nums = [10, 20, 70, 40, 50, 60]

As we go through nums, we will aggregate the sums of ascending subarrays and keep track of the maximum sum encountered. We will utilize two variables, t and ans, throughout the process. Follow these steps:

Initialize t with the first element and ans with 0, since we have not encountered any subarrays yet. t = 10, ans = 0
Move to the next element (20) and compare it with the previous one (10). Since 20 is greater than 10, it continues an ascending subarray. Add 20 to t and update ans. t = 10 + 20 = 30, ans = max(0, 30) = 30
The next element is 70, which is greater than 20, so add 70 to t and update ans. t = 30 + 70 = 100, ans = max(30, 100) = 100
We now encounter 40. Since 40 is not greater than the preceding element 70, the ascending condition is broken. Here, we start a new subarray. Thus, reset t to the value of the current element, 40, and ans remains the same as it's still the maximum sum encountered. t = 40, ans = 100
The next element is 50, which is greater than 40. We add 50 to t and compare it with ans. t = 40 + 50 = 90, ans remains 100 as 90 is not greater than 100.
Lastly, 60 is greater than 50, so add it to t and check against ans. t = 90 + 60 = 150, ans = max(100, 150) = 150

At the end of this process, ans holds the value of the highest sum of an ascending subarray within the array nums, which in this case is 150. This sum comes from the subarray [40, 50, 60].

By following this iterative approach, we find the maximum sum efficiently with just one pass through the input array.

Solution Implementation

1# The Solution class encapsulates the algorithm to find the maximum ascending subarray sum.
2class Solution:
3    def maxAscendingSum(self, nums: List[int]) -> int:
4        max_sum = temp_sum = nums[0]  # Initialize max_sum and temp_sum with the first element
5      
6        # Iterate through the list starting from the second element
7        for i in range(1, len(nums)):
8            # If the current element is greater than the previous element, add it to the temp_sum.
9            if nums[i] > nums[i - 1]:
10                temp_sum += nums[i]
11            else:
12                # Else, assign current element to temp_sum as the start of a new subarray.
13                temp_sum = nums[i]
14
15            # Update max_sum if temp_sum is greater.
16            max_sum = max(max_sum, temp_sum)
17      
18        return max_sum  # Return the maximum sum found.
19

1class Solution {
2    public int maxAscendingSum(int[] nums) {
3        int maxSum = 0; // This variable will store the maximum ascending subarray sum
4        int currentSum = 0; // This variable will keep the current subarray sum
5      
6        // Iterate over all the elements in the array
7        for (int i = 0; i < nums.length; ++i) {
8            // Check if the current element is greater than the previous element or it is the first element
9            if (i == 0 || nums[i] > nums[i - 1]) {
10                currentSum += nums[i]; // Add the current element to the current sum
11                maxSum = Math.max(maxSum, currentSum); // Update the maxSum with the larger of the two sums
12            } else {
13                currentSum = nums[i]; // If the current element is not greater then start a new subarray from the current element
14            }
15        }
16        return maxSum; // Return the maximum sum of ascending subarray found
17    }
18}
19

1class Solution {
2public:
3    // Function to find the maximum ascending subarray sum.
4    int maxAscendingSum(vector<int>& nums) {
5        int maxSum = 0; // Variable to store the maximum sum of ascending subarray
6        int currentSum = 0; // Variable to store the current subarray sum
7
8        // Iterate through all the elements of the input vector
9        for (int i = 0; i < nums.size(); ++i) {
10            // Start a new subarray if we are at the first element or if the current element
11            // is greater than the previous one, thus obeying the ascending order
12            if (i == 0 || nums[i] > nums[i - 1]) {
13                currentSum += nums[i]; // Accumulate current subarray sum
14                maxSum = max(maxSum, currentSum); // Update maxSum if currentSum is greater
15            } else {
16                // If the current element is not greater than the previous one,
17                // start a new subarray sum from the current element
18                currentSum = nums[i];
19            }
20        }
21
22        // Return the maximum sum of ascending subarray found
23        return maxSum;
24    }
25};
26

1/**
2 * Calculates the maximum ascending subarray sum in an array of numbers
3 * @param nums - The given array of numbers
4 * @returns The maximum sum of an ascending subarray
5 */
6function maxAscendingSum(nums: number[]): number {
7    const length = nums.length; // Length of the input array
8    let maxSum = nums[0]; // Initialize maxSum as the first element
9    let currentSum = nums[0]; // Initialize currentSum as the first element
10
11    // Iterate through the array starting from the second element
12    for (let i = 1; i < length; i++) {
13        // If the current element is not larger than the previous one,
14        // compare and update the maxSum with the currentSum so far,
15        // then reset the currentSum to the current element
16        if (nums[i] <= nums[i - 1]) {
17            maxSum = Math.max(maxSum, currentSum);
18            currentSum = nums[i];
19        } else {
20            // If the current element is larger, add it to the currentSum
21            currentSum += nums[i];
22        }
23    }
24
25    // Return the maximum sum between maxSum and the currentSum
26    // to cover the case where the last element was part of the ascending sequence
27    return Math.max(maxSum, currentSum);
28}
29

Time and Space Complexity

The time complexity of the given code is O(n), where n is the length of the nums list. This is because the code iterates through the nums list once, with each operation within the loop having a constant time complexity.

The space complexity of the code is O(1) as it uses a fixed amount of extra space; only two integer variables ans and t are used, regardless of the input size.

Learn more about how to find time and space complexity quickly using problem constraints.

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