Sum of Subarray Ranges

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ArraysStacks

You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.

Return the sum of all subarray ranges of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0 
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.

Example 2:

Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.

Example 3:

Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.

Constraints:

1 <= nums.length <= 1000
-10⁹ <= nums[i] <= 10⁹

Follow-up: Could you find a solution with O(n) time complexity?

Try coding it in LeetCode

Sum of Subarray Ranges

Medium

Asked by:

34 views

Topics:

ArraysStacks

You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.

Return the sum of all subarray ranges of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0 
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.

Example 2:

Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.

Example 3:

Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.

Constraints:

1 <= nums.length <= 1000
-10⁹ <= nums[i] <= 10⁹

Follow-up: Could you find a solution with O(n) time complexity?

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 `nums` array and the range of values within the array?
Can the input array `nums` contain negative numbers or zero?
Are there any specific edge cases I should consider, such as an empty input array or an array with only one element?
Are duplicate numbers allowed in the input array `nums`, and if so, how should they be handled when calculating subarray ranges?
Could you provide an example of the calculation for the sum of subarray ranges with a small sample input?

Brute Force Solution

Approach

The brute force way to find the sum of subarray ranges means we'll look at every possible group of numbers in our list. For each group, we'll find the biggest and smallest numbers, then subtract the smallest from the biggest. Finally, we'll add up all those differences to get our final answer.

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

First, consider groups that have only one number.
Then, consider groups that have two numbers, taking each pair one by one.
Next, consider groups that have three numbers, then four, and so on, until you've considered the group that contains all the numbers in the list.
For each group you look at, identify the largest and smallest numbers in that group.
Calculate the difference between the largest and smallest number in that group.
Add this difference to a running total.
Once you've gone through every possible group, the final total is your answer.

Code Implementation

def sub_array_ranges_brute_force(numbers):
    result = 0

    # Iterate through all possible starting indices
    for start_index in range(len(numbers)):

        # Iterate through all possible ending indices
        for end_index in range(start_index, len(numbers)):

            # Initialize min/max with the starting element.
            minimum_number = numbers[start_index]
            maximum_number = numbers[start_index]

            # Iterate through current subarray
            for current_index in range(start_index, end_index + 1):

                # Find the min and max values in the subarray
                minimum_number = min(minimum_number, numbers[current_index])
                maximum_number = max(maximum_number, numbers[current_index])

            # Accumulate difference to final result
            result += maximum_number - minimum_number

    return result

Big(O) Analysis

Time Complexity

O(n³) – The provided approach iterates through all possible subarrays. There are roughly n(n+1)/2 possible subarrays in an array of size n, which simplifies to approximately n²/2 subarrays. For each subarray, the algorithm finds the minimum and maximum elements, which takes O(n) time in the worst case (specifically, it can be at most the length of the subarray which will be <= n). Thus, the total time complexity is (n²/2) * n, which simplifies to O(n³).

Space Complexity

O(1) – The provided brute force algorithm calculates the range (max - min) for each subarray and sums these ranges. It does not create any auxiliary data structures whose size scales with the input size N (the number of elements in the input list). It only uses a few constant space variables to store the current minimum, maximum, and the running sum of subarray ranges. Thus, the space complexity is constant.

Optimal Solution

Approach

Instead of checking the range of every possible group, the key idea is to find how many times each individual number acts as the smallest and largest within some group. Then we use this information to directly calculate the sum of ranges, avoiding redundant calculations. This dramatically improves efficiency.

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

For each number in the group, figure out how many groups exist where it is the smallest number.
Also, for each number in the group, figure out how many groups exist where it is the largest number.
Multiply each number by the number of times it's the largest, and add all of those results together.
Multiply each number by the number of times it's the smallest, and add all of those results together.
Subtract the total of the 'smallest' products from the total of the 'largest' products. The result is the answer you are looking for.

Code Implementation

def sub_array_ranges(group):
    group_length = len(group)
    max_sum = 0
    min_sum = 0

    # Calculate contribution of each num as max.
    for i in range(group_length):
        left_boundary = 1
        right_boundary = 1
        while i - left_boundary >= 0 and group[i - left_boundary] <= group[i]:
            left_boundary += 1
        while i + right_boundary < group_length and group[i + right_boundary] < group[i]:
            right_boundary += 1

        # Accumulate how each number contributes as largest.
        max_sum += group[i] * left_boundary * right_boundary

    # Calculate contribution of each num as min.
    for i in range(group_length):
        left_boundary = 1
        right_boundary = 1
        while i - left_boundary >= 0 and group[i - left_boundary] >= group[i]:
            left_boundary += 1
        while i + right_boundary < group_length and group[i + right_boundary] > group[i]:
            right_boundary += 1

        # Accumulate how each number contributes as smallest.
        min_sum += group[i] * left_boundary * right_boundary

    # Difference between max and min contributions gives the result.
    return max_sum - min_sum

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each element of the input array once to determine how many times that element is the minimum and maximum value within subarrays. For each element, it effectively expands outwards to determine the boundaries of the subarrays where it holds the minimum or maximum value. This expansion involves comparing the current element with elements to its left and right. However, each element is visited a limited number of times that is relative to the input array size, as the algorithm needs to find the nearest smaller and bigger element to the left and right respectively to determine the contribution of the number in the total range. The number of operations are therefore proportional to the size of the array. Thus, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm needs to determine the number of times each number acts as the smallest and largest within a subarray. While the description doesn't explicitly mention a data structure, to efficiently calculate these counts, it implicitly requires storing information about the boundaries of subarrays where each number is the minimum or maximum. This can be achieved using stacks to track the next smaller/larger element on both sides for each number, where the stacks can grow up to size N in the worst case. Therefore, auxiliary space proportional to the size of the input array, N, is needed.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there are no subarrays to calculate ranges from.
Array with a single element	Return 0 as a single-element subarray has a range of 0 (element - element).
Array with all identical values	The sum of subarray ranges will be 0 because each subarray's minimum and maximum are the same.
Array with negative numbers	The algorithm should correctly handle negative numbers when calculating the min and max of each subarray.
Array with large numbers that might lead to integer overflow when calculating the sum of ranges.	Use a 64-bit integer type (long in Java/C++) to store the sum of ranges to prevent overflow.
Very large input array (performance)	A brute-force O(n^2) solution might be too slow; consider using stack-based or segment tree approaches for O(n) or O(n log n) performance.
Array containing Integer.MAX_VALUE and Integer.MIN_VALUE	Ensure calculations handle the boundaries correctly when computing min/max and their differences.
The array has many subarrays with the same range	The algorithm should still process each subarray individually to calculate the sum, regardless of repeated range values.

Null or empty input array

How to Handle:

Return 0 immediately as there are no subarrays to calculate ranges from.

Array with a single element

How to Handle:

Return 0 as a single-element subarray has a range of 0 (element - element).

Array with all identical values

How to Handle:

The sum of subarray ranges will be 0 because each subarray's minimum and maximum are the same.

Array with negative numbers

How to Handle:

The algorithm should correctly handle negative numbers when calculating the min and max of each subarray.

Array with large numbers that might lead to integer overflow when calculating the sum of ranges.

How to Handle:

Use a 64-bit integer type (long in Java/C++) to store the sum of ranges to prevent overflow.

Very large input array (performance)

How to Handle:

A brute-force O(n^2) solution might be too slow; consider using stack-based or segment tree approaches for O(n) or O(n log n) performance.

Array containing Integer.MAX_VALUE and Integer.MIN_VALUE

How to Handle:

Ensure calculations handle the boundaries correctly when computing min/max and their differences.

The array has many subarrays with the same range

How to Handle:

The algorithm should still process each subarray individually to calculate the sum, regardless of repeated range values.