Range Sum of Sorted Subarray Sums

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ArraysBinary SearchGreedy Algorithms

You are given the array nums consisting of n positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.

Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13 
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.

Example 2:

Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.

Example 3:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50

Constraints:

n == nums.length
1 <= nums.length <= 1000
1 <= nums[i] <= 100
1 <= left <= right <= n * (n + 1) / 2

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 constraints on the size of the input array `nums` (i.e., the maximum length)?
Can the elements in the input array `nums` be negative, zero, or only positive?
Could you clarify the definition of 'sorted subarray sums'? Specifically, are we sorting the individual subarrays before summing, or are we sorting the sums of all possible subarrays?
Are the `left` and `right` indices 1-based or 0-based?
What should I return if `left` is greater than `right`?

Brute Force Solution

Approach

The problem wants us to find the sum of a specific range of sorted subarray sums. The brute force approach is all about calculating every possible subarray sum, putting them all together, sorting them, and then summing the ones in the desired range. It's like figuring out every possible combination and then filtering to what you need.

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

First, find all possible subarrays from the given set of numbers.
For each of these subarrays, calculate the sum of its numbers.
Now, gather up all these sums that you calculated.
Next, arrange these sums in ascending order, from smallest to largest.
Identify the sums within the specified range, starting from a particular position and ending at another.
Finally, add up only those sums that fall within the chosen range to get the final answer.

Code Implementation

def range_sum_sorted_subarray_sums_brute_force(numbers, number_of_elements, left_index, right_index):
    subarray_sums = []

    # Generate all possible subarray sums
    for start_index in range(number_of_elements):
        for end_index in range(start_index, number_of_elements):
            current_subarray_sum = 0
            for index in range(start_index, end_index + 1):
                current_subarray_sum += numbers[index]
            subarray_sums.append(current_subarray_sum)

    # Sort the subarray sums in ascending order
    subarray_sums.sort()

    total_sum_in_range = 0

    # Iterate through the sorted subarray sums and sum the elements in the specified range
    # Need to adjust the indices to be 0-based
    for index in range(left_index - 1, right_index):
        total_sum_in_range += subarray_sums[index]

    return total_sum_in_range

Big(O) Analysis

Time Complexity

O(n² log n) – The brute force approach involves generating all subarray sums, which takes O(n²) time because we iterate through all possible start and end indices of subarrays. We then sort these O(n²) subarray sums which requires O(n² log n) time. Finally, summing a range of elements in the sorted array takes O(n) time, but this is dominated by the sorting step. Therefore the overall time complexity is O(n² log n).

Space Complexity

O(N^2) – The described algorithm calculates all possible subarray sums. In the worst-case scenario, there are N(N+1)/2 possible subarrays, where N is the size of the input array. These sums are stored in a list before being sorted, resulting in auxiliary space proportional to the number of subarrays. Therefore, the space complexity is O(N^2).

Optimal Solution

Approach

The key idea is to avoid calculating every single subarray sum directly. We can efficiently compute the sums of all subarrays by cleverly accumulating sums and sorting them. This allows us to quickly find the range sum we need.

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

First, calculate the sum of all possible subarrays and store these sums.
Think of it as building a list of sums, where each sum represents a continuous section of the original data.
Then, sort all these subarray sums in ascending order.
Next, determine the specific section of the sorted list of sums that contributes to our final answer based on the given range.
Finally, add up all the sorted subarray sums within that identified section. This is our answer.

Code Implementation

def range_sum_sorted_subarray_sums(nums, n, left, right):
    subarray_sums = []
    for i in range(n):
        current_sum = 0
        for j in range(i, n):
            current_sum += nums[j]
            subarray_sums.append(current_sum)

    # Sorting is crucial for efficiently calculating the range sum
    subarray_sums.sort()

    total_sum = 0
    # Adjust indices to be 0-based.
    left -= 1
    right -= 1

    # Iterate through the desired range of sorted subarray sums.
    for i in range(left, right + 1):
        total_sum += subarray_sums[i]

    # Apply modulo to prevent overflow, as problem constraints dictate.
    return total_sum % (10**9 + 7)

Big(O) Analysis

Time Complexity

O(n log n) – Calculating all subarray sums involves iterating through all possible subarrays, resulting in approximately n(n+1)/2 sums. Storing these sums in a list takes O(n^2) space. The most significant operation is sorting this list of subarray sums, which has a length on the order of n^2. Sorting n^2 elements takes O(n^2 log(n^2)) time, which simplifies to O(n^2 log n). After the sorting, calculating the final range sum takes O(n^2) which is dominated by the sorting. Thus the overall complexity is O(n^2 log n).

Space Complexity

O(N^2) – The algorithm first calculates the sum of all possible subarrays and stores these sums. In the worst case, the number of subarrays for an array of size N is N*(N+1)/2, which is on the order of N^2. These sums are stored in a list, resulting in auxiliary space proportional to the number of subarrays. Sorting the subarray sums also happens in-place, and its space complexity should be lower than the auxiliary list. Therefore, the dominant space complexity is O(N^2) due to storing the subarray sums.

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 since no subarray sums exist in an empty array.
Array with a single element (nums has length 1)	Calculate and return the single subarray sum which equals the element itself if left==1 and right==1, otherwise return 0.
Large array size (n is close to its maximum constraint)	Ensure the solution's time complexity is efficient (e.g., O(n log n)) to avoid exceeding time limits when calculating all subarray sums and sorting them.
Large range (left and right are close to maximum allowed)	Calculate the subarray sums lazily and only when the index is within the range [left, right] to avoid unnecessary calculations.
Array containing large positive numbers that may lead to integer overflow when calculating subarray sums.	Use 64-bit integers (long) to store subarray sums to prevent overflow.
Input array with all elements equal to zero	The sorted subarray sums will all be zero, requiring correct handling of range calculations with multiple zeros.
The sum of the subarray sums exceeds the maximum integer value before modulo.	Apply the modulo operation (%) after each addition to prevent overflow during the sum calculation of the selected sorted subarray sums.
left > right	If left is greater than right, return 0 immediately, as it's an invalid range request.

Empty input array (nums is null or has length 0)

How to Handle:

Return 0 since no subarray sums exist in an empty array.

Array with a single element (nums has length 1)

How to Handle:

Calculate and return the single subarray sum which equals the element itself if left==1 and right==1, otherwise return 0.

Large array size (n is close to its maximum constraint)

How to Handle:

Ensure the solution's time complexity is efficient (e.g., O(n log n)) to avoid exceeding time limits when calculating all subarray sums and sorting them.

Large range (left and right are close to maximum allowed)

How to Handle:

Calculate the subarray sums lazily and only when the index is within the range [left, right] to avoid unnecessary calculations.

Array containing large positive numbers that may lead to integer overflow when calculating subarray sums.

How to Handle:

Use 64-bit integers (long) to store subarray sums to prevent overflow.

Input array with all elements equal to zero

How to Handle:

The sorted subarray sums will all be zero, requiring correct handling of range calculations with multiple zeros.

The sum of the subarray sums exceeds the maximum integer value before modulo.

How to Handle:

Apply the modulo operation (%) after each addition to prevent overflow during the sum calculation of the selected sorted subarray sums.

left > right

How to Handle:

If left is greater than right, return 0 immediately, as it's an invalid range request.