Problem Description

You are given an array nums containing n positive integers. Your task involves creating a new array by computing sums of all possible continuous subarrays and then finding a specific range sum.

Here's what you need to do:

1. Generate all continuous subarray sums: For every possible continuous subarray in nums, calculate its sum. This includes:

   • Single elements: nums[0], nums[1], ..., nums[n-1]
   • Two-element subarrays: nums[0]+nums[1], nums[1]+nums[2], ...
   • Three-element subarrays: nums[0]+nums[1]+nums[2], ...
   • And so on, up to the sum of the entire array

2. Sort these sums: Take all these subarray sums and sort them in non-decreasing order. This creates a new sorted array with n * (n + 1) / 2 elements (since that's how many continuous subarrays exist).

3. Find the range sum: Given two indices left and right (both 1-indexed), calculate the sum of all elements from position left to position right in the sorted array.

4. Return the result modulo 10^9 + 7: Since the answer might be very large, return it modulo 10^9 + 7.

For example, if nums = [1, 2, 3, 4]:

• The continuous subarray sums would be: [1, 2, 3, 4, 3, 5, 7, 6, 9, 10] (representing sums of [1], [2], [3], [4], [1,2], [2,3], [3,4], [1,2,3], [2,3,4], [1,2,3,4])
• After sorting: [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]
• If left = 1 and right = 5, the answer would be 1 + 2 + 3 + 3 + 4 = 13

Intuition

The problem asks us to work with subarray sums in a sorted manner. The most straightforward way to think about this is to break it down into manageable steps.

First, we need to generate all possible subarray sums. For an array of size n, we know there are exactly n * (n + 1) / 2 subarrays. Why? Because from each starting position i, we can form (n - i) subarrays ending at different positions. This gives us n + (n-1) + (n-2) + ... + 1 = n * (n + 1) / 2 total subarrays.

To generate these sums, we can use a nested loop approach. The outer loop fixes the starting position, and the inner loop extends the subarray one element at a time. As we extend, we can maintain a running sum rather than recalculating from scratch each time. This is efficient because adding one more element to a subarray just means adding that element to the previous sum.

Once we have all the subarray sums, sorting them is straightforward - we just use a standard sorting algorithm. The key insight here is that we don't need to maintain information about which subarray produced which sum; we only care about the values themselves.

Finally, to get the sum of elements from index left to right (1-indexed), we convert to 0-indexed by subtracting 1, then sum up the elements in that range of our sorted array.

The modulo operation 10^9 + 7 is applied at the end to prevent integer overflow, which is a common requirement in competitive programming when dealing with potentially large sums.

This brute force approach works well for the given constraints because generating and sorting n * (n + 1) / 2 elements is computationally feasible for reasonable values of n.

Learn more about Two Pointers, Binary Search, Prefix Sum and Sorting patterns.

Solution Approach

The implementation follows a simulation approach where we generate all subarray sums, sort them, and then calculate the required range sum.

Step 1: Generate all subarray sums

We use a nested loop structure to generate all possible subarray sums:

• The outer loop with index i iterates from 0 to n-1, representing the starting position of each subarray
• For each starting position i, we initialize a sum variable s = 0
• The inner loop with index j iterates from i to n-1, representing the ending position of each subarray
• As we extend the subarray by including nums[j], we update the running sum: s += nums[j]
• Each computed sum s is appended to our result array arr

This approach efficiently computes all subarray sums in O(n^2) time by maintaining a running sum instead of recalculating from scratch for each subarray.

Step 2: Sort the array

Once we have all n * (n + 1) / 2 subarray sums in arr, we sort them using Python's built-in sort() method, which typically uses Timsort with O(k log k) complexity where k = n * (n + 1) / 2.

Step 3: Calculate the range sum

Since the problem uses 1-based indexing while Python uses 0-based indexing, we need to adjust:

• The element at position left (1-indexed) corresponds to index left - 1 in our array
• The element at position right (1-indexed) corresponds to index right - 1 in our array

We use Python's slicing notation arr[left - 1 : right] to extract elements from index left - 1 to right - 1 inclusive (note that the end index in Python slicing is exclusive).

Step 4: Apply modulo

Finally, we sum all elements in the extracted range and apply modulo 10^9 + 7 to prevent overflow and meet the problem's requirements.

The overall time complexity is O(n^2 log n) dominated by the sorting step, and the space complexity is O(n^2) for storing all subarray sums.

Example Walkthrough

Let's walk through a small example with nums = [1, 2, 3], left = 2, and right = 5.

Step 1: Generate all subarray sums

We iterate through all possible subarrays:

• Start at index 0 (value 1):
  • Subarray [1]: sum = 1
  • Subarray [1,2]: sum = 1+2 = 3
  • Subarray [1,2,3]: sum = 1+2+3 = 6
• Start at index 1 (value 2):
  • Subarray [2]: sum = 2
  • Subarray [2,3]: sum = 2+3 = 5
• Start at index 2 (value 3):
  • Subarray [3]: sum = 3

This gives us the array of sums: [1, 3, 6, 2, 5, 3]

Step 2: Sort the array

Sorting [1, 3, 6, 2, 5, 3] gives us: [1, 2, 3, 3, 5, 6]

Step 3: Calculate the range sum

With left = 2 and right = 5 (1-indexed):

• Convert to 0-indexed: we need elements from index 1 to index 4
• Extract elements: arr[1:5] = [2, 3, 3, 5]
• Sum these elements: 2 + 3 + 3 + 5 = 13

Step 4: Apply modulo

13 % (10^9 + 7) = 13

The final answer is 13.

This example demonstrates how the solution systematically generates all possible subarray sums using nested loops with a running sum optimization, sorts them to create an ordered list, and then extracts the required range to compute the final result.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n² × log n)

The time complexity breaks down as follows:

• The nested loops (lines 4-8) iterate through all possible subarrays. The outer loop runs n times, and for each iteration i, the inner loop runs (n - i) times. This generates a total of n + (n-1) + (n-2) + ... + 1 = n(n+1)/2 subarrays, which is O(n²) operations.
• Each subarray sum is appended to the array, taking O(1) time per append.
• After generating all subarray sums, the array is sorted (line 9), which takes O(k × log k) where k is the number of elements. Since we have n(n+1)/2 elements, this is O(n² × log n²) = O(n² × 2log n) = O(n² × log n).
• The final sum operation (line 11) takes at most O(n²) time in the worst case when summing all elements.

The dominant operation is the sorting step, giving us a total time complexity of O(n² × log n).

Space Complexity: O(n²)

The space complexity is determined by:

• The arr list stores all possible subarray sums. With n(n+1)/2 subarrays, this requires O(n²) space.
• The sorting operation may use additional O(log n) to O(n²) space depending on the implementation, but this doesn't exceed O(n²).
• Other variables (s, i, j, mod) use constant O(1) space.

Therefore, the overall space complexity is O(n²).

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

Common Pitfalls

1. Forgetting to Apply Modulo Operation

One of the most common mistakes is forgetting to apply the modulo operation or applying it incorrectly. The problem explicitly requires the result to be returned modulo 10^9 + 7.

Incorrect approach:

# Forgetting modulo entirely
result = sum(subarray_sums[left - 1:right])
return result

# Or applying modulo too early (though less critical in Python)
for i in range(n):
    current_sum = 0
    for j in range(i, n):
        current_sum = (current_sum + nums[j]) % MOD  # Unnecessary here
        subarray_sums.append(current_sum)

Why it matters: For large arrays with large values, the sum can exceed the maximum integer limit in some languages. While Python handles big integers automatically, the problem requires modulo for consistency and to match expected output format.

Solution: Always apply modulo to the final result as specified in the problem.

2. Index Confusion Between 0-based and 1-based

The problem uses 1-based indexing for left and right, but Python uses 0-based indexing. This is a frequent source of errors.

Incorrect approaches:

# Using indices directly without conversion
result = sum(subarray_sums[left:right]) % MOD  # Wrong!

# Off-by-one error in slicing
result = sum(subarray_sums[left:right + 1]) % MOD  # Wrong!

Why it matters: These mistakes will either include wrong elements or miss required elements, leading to incorrect results.

Solution: Remember that for 1-based indices:

• Position left maps to index left - 1
• Position right maps to index right - 1
• Python slicing [start:end] includes start but excludes end, so use [left-1:right]

3. Inefficient Subarray Sum Calculation

A naive approach might recalculate the entire sum for each subarray from scratch:

Inefficient approach:

subarray_sums = []
for i in range(n):
    for j in range(i, n):
        # Recalculating sum from scratch each time
        subarray_sum = sum(nums[i:j+1])  # O(n) operation inside nested loops!
        subarray_sums.append(subarray_sum)

Why it matters: This adds an extra O(n) factor to the time complexity, making it O(n³) instead of O(n²), which can cause timeouts for large inputs.

Solution: Use a running sum approach as shown in the correct implementation, where you maintain and update the sum incrementally as you extend the subarray.

4. Memory Issues with Very Large Arrays

While less common, for extremely large values of n, storing all O(n²) subarray sums might cause memory issues.

Potential issue:

# For n = 10,000, this creates an array with 50,005,000 elements
subarray_sums = []  # Could consume significant memory

Solution: For most competitive programming problems, this is acceptable. However, if memory becomes an issue, consider:

• Using a generator-based approach if only specific ranges are needed
• Implementing a more space-efficient algorithm using heaps for finding k-th smallest elements
• Note that for this specific problem, we need all sums sorted, so the space usage is generally unavoidable