Maximum Sum Score of Array

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Arrays

You are given a 0-indexed integer array nums of length n.

The sum score of nums at an index i is defined as the maximum of the sum of the first i + 1 elements and the sum of the last n - i elements.

Return the maximum sum score of nums at any index.

Example 1:

Input: nums = [4,3,2,1]
Output: 10
Explanation:
- The sum score at index 0 is max(4, 4+3+2+1) = max(4, 10) = 10.
- The sum score at index 1 is max(4+3, 3+2+1) = max(7, 6) = 7.
- The sum score at index 2 is max(4+3+2, 2+1) = max(9, 3) = 9.
- The sum score at index 3 is max(4+3+2+1, 1) = max(10, 1) = 10.
So, the maximum sum score of nums at any index is 10.

Example 2:

Input: nums = [7,9,5,8,1,3]
Output: 31
Explanation:
- The sum score at index 0 is max(7, 7+9+5+8+1+3) = max(7, 33) = 33.
- The sum score at index 1 is max(7+9, 9+5+8+1+3) = max(16, 26) = 26.
- The sum score at index 2 is max(7+9+5, 5+8+1+3) = max(21, 17) = 21.
- The sum score at index 3 is max(7+9+5+8, 8+1+3) = max(29, 12) = 29.
- The sum score at index 4 is max(7+9+5+8+1, 1+3) = max(30, 4) = 30.
- The sum score at index 5 is max(7+9+5+8+1+3, 3) = max(33, 3) = 33.
So, the maximum sum score of nums at any index is 33.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the integer values within the array?
Can the input array be empty or null?
If there are multiple pairs that result in the maximum sum score, is any one of them acceptable, or is there a specific pair that should be preferred based on a tie-breaking criteria?
Is the array guaranteed to contain at least two elements?
Could you define 'sum score' with some examples to make sure I understand correctly?

Brute Force Solution

Approach

The brute force method for this problem involves trying out every single possible combination to find the maximum sum. We'll explore each way to split the array and calculate the score for each split.

Here's how the algorithm would work step-by-step:

Consider all possible starting points for a "left part" of the array. This means trying a left part with only the first number, then the first two numbers, then the first three, and so on until the entire array is the "left part".
For each of those "left part" choices, the remaining part of the array becomes the "right part".
Calculate the sum of the numbers in the "left part" and the sum of the numbers in the "right part".
Determine the score for this particular split by comparing the "left part" sum and the "right part" sum; whichever is larger becomes the score.
Record the score for this arrangement.
Repeat the process by trying all possible arrangements where the "left part" and "right part" are split differently.
After exploring every possible split, compare all the recorded scores and select the highest one. This highest score is the maximum sum score.

Code Implementation

def maximum_sum_score_of_array_brute_force(numbers):
    maximum_score = float('-inf')

    for left_partition_end_index in range(1, len(numbers) + 1):
        # Iterate through all possible ending indices for the left partition
        left_partition = numbers[:left_partition_end_index]
        right_partition = numbers[left_partition_end_index:]

        left_partition_sum = sum(left_partition)
        right_partition_sum = sum(right_partition)

        # Calculate the score as the maximum of the two partition sums
        current_score = max(left_partition_sum, right_partition_sum)
        
        # Update the maximum score if the current score is higher
        maximum_score = max(maximum_score, current_score)
    
    return maximum_score

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible starting points for the left part of the array. For each starting point, it calculates the sums of both the left and right parts. Since we are considering all possible left parts, ranging from a single element to the entire array, the outer loop runs 'n' times. Inside this loop, the sums for left and right parts are calculated which takes O(n) time. Therefore, the total time complexity is dominated by the nested iterations, resulting in approximately n * n operations, which simplifies to O(n²).

Space Complexity

O(1) – The brute force method calculates left and right sums within the loops, storing these sums in temporary variables each time. No auxiliary data structures that scale with the input size N (the length of the array) are used. Therefore, the space complexity is constant, irrespective of the input array's size. We only use a fixed number of variables to store intermediate calculations.

Optimal Solution

Approach

The challenge asks us to find the largest possible 'score' obtainable from an array. We calculate the score by comparing the cumulative sum from the left against the cumulative sum from the right for each position in the array. The key is to compute and compare these sums efficiently to identify the position with the maximum score.

Here's how the algorithm would work step-by-step:

First, calculate the total sum of all the numbers in the given list.
Next, imagine you're walking through the list from the start. Keep a running total of the numbers you've seen so far.
At each number, figure out the sum of all the numbers to the left (the running total) and compare it to the sum of all numbers to the right (the total sum minus the running total and the current number).
Choose the larger of these two sums (left or right) - this is your 'score' for that position.
Remember the highest 'score' you've seen so far as you walk through the list.
After checking every position in the list, the highest score you remembered is the final answer.

Code Implementation

def maximum_sum_score(numbers):
    total_sum = sum(numbers)
    running_sum = 0
    maximum_score = 0

    for index in range(len(numbers)):
        # Update the running sum with the current number.
        running_sum += numbers[index]

        # Calculate the left and right sums.
        left_sum = running_sum
        right_sum = total_sum - running_sum

        # The score at current index is the maximum of left and right sums.
        current_score = max(left_sum, right_sum)

        # Keep track of the maximum score encountered so far.
        if current_score > maximum_score:
            maximum_score = current_score

    # After iterating through the array return the maximum score.
    return maximum_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm first calculates the total sum of the array, which takes O(n) time. Then, it iterates through the array once. Inside the loop, it maintains a running sum from the left. For each element, it calculates the sum from the left (running sum) and the sum from the right (total sum minus the running sum and the current element), both of which are O(1) operations. Since the loop iterates n times, the overall time complexity is dominated by the loop, making it O(n).

Space Complexity

O(1) – The algorithm calculates the total sum and maintains a running sum. It then iterates through the array, comparing the left and right sums at each position, without using any auxiliary data structures that scale with the input size N (where N is the number of elements in the input array). Only a few constant-size variables are used to store the total sum, running sum, and maximum score. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null input array	Throw an IllegalArgumentException or return 0 after checking for null input.
Empty input array	Return 0 since there are no elements to sum.
Array with a single element	Return the single element's value as both prefix and suffix sums are equal to it.
Array with all negative numbers	Calculate prefix and suffix sums correctly, ensuring negative sums are handled.
Array with all zeros	Return 0 as both prefix and suffix sums are 0 for all elements.
Array with very large positive numbers leading to potential integer overflow	Use long data type to store prefix and suffix sums to prevent integer overflow.
Array with extreme differences in values (e.g., large positive and large negative numbers)	Ensure calculation of prefix/suffix sums and maximum scores is accurate, handling both positive and negative values.
Maximum sized array causing memory issues	The iterative prefix/suffix sum approach is efficient with O(n) space and time complexity and avoids excessive memory usage.

Null input array

How to Handle:

Throw an IllegalArgumentException or return 0 after checking for null input.

Empty input array

How to Handle:

Return 0 since there are no elements to sum.

Array with a single element

How to Handle:

Return the single element's value as both prefix and suffix sums are equal to it.

Array with all negative numbers

How to Handle:

Calculate prefix and suffix sums correctly, ensuring negative sums are handled.

Array with all zeros

How to Handle:

Return 0 as both prefix and suffix sums are 0 for all elements.

Array with very large positive numbers leading to potential integer overflow

How to Handle:

Use long data type to store prefix and suffix sums to prevent integer overflow.

Array with extreme differences in values (e.g., large positive and large negative numbers)

How to Handle:

Ensure calculation of prefix/suffix sums and maximum scores is accurate, handling both positive and negative values.

Maximum sized array causing memory issues

How to Handle:

The iterative prefix/suffix sum approach is efficient with O(n) space and time complexity and avoids excessive memory usage.