Sum of Imbalance Numbers of All Subarrays

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Arrays

The imbalance number of a 0-indexed integer array arr of length n is defined as the number of indices in sarr = sorted(arr) such that:

0 <= i < n - 1, and
sarr[i+1] - sarr[i] > 1

Here, sorted(arr) is the function that returns the sorted version of arr.

Given a 0-indexed integer array nums, return the sum of imbalance numbers of all its subarrays.

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

Example 1:

Input: nums = [2,3,1,4]
Output: 3
Explanation: There are 3 subarrays with non-zero imbalance numbers:
- Subarray [3, 1] with an imbalance number of 1.
- Subarray [3, 1, 4] with an imbalance number of 1.
- Subarray [1, 4] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3.

Example 2:

Input: nums = [1,3,3,3,5]
Output: 8
Explanation: There are 7 subarrays with non-zero imbalance numbers:
- Subarray [1, 3] with an imbalance number of 1.
- Subarray [1, 3, 3] with an imbalance number of 1.
- Subarray [1, 3, 3, 3] with an imbalance number of 1.
- Subarray [1, 3, 3, 3, 5] with an imbalance number of 2. 
- Subarray [3, 3, 3, 5] with an imbalance number of 1. 
- Subarray [3, 3, 5] with an imbalance number of 1.
- Subarray [3, 5] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 8.

Constraints:

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

Try coding it in LeetCode

Sum of Imbalance Numbers of All Subarrays

Hard

Asked by:

13 views

Topics:

Arrays

The imbalance number of a 0-indexed integer array arr of length n is defined as the number of indices in sarr = sorted(arr) such that:

0 <= i < n - 1, and
sarr[i+1] - sarr[i] > 1

Here, sorted(arr) is the function that returns the sorted version of arr.

Given a 0-indexed integer array nums, return the sum of imbalance numbers of all its subarrays.

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

Example 1:

Input: nums = [2,3,1,4]
Output: 3
Explanation: There are 3 subarrays with non-zero imbalance numbers:
- Subarray [3, 1] with an imbalance number of 1.
- Subarray [3, 1, 4] with an imbalance number of 1.
- Subarray [1, 4] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3.

Example 2:

Input: nums = [1,3,3,3,5]
Output: 8
Explanation: There are 7 subarrays with non-zero imbalance numbers:
- Subarray [1, 3] with an imbalance number of 1.
- Subarray [1, 3, 3] with an imbalance number of 1.
- Subarray [1, 3, 3, 3] with an imbalance number of 1.
- Subarray [1, 3, 3, 3, 5] with an imbalance number of 2. 
- Subarray [3, 3, 3, 5] with an imbalance number of 1. 
- Subarray [3, 3, 5] with an imbalance number of 1.
- Subarray [3, 5] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 8.

Constraints:

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

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 is the expected data type of the input array elements, and what is the range of values for each element?
Can the input array be empty or contain only a single element? If so, what should the function return in these cases?
Could you please clarify what constitutes the 'imbalance number' of a subarray with a concrete example, especially regarding consecutive numbers or repeated values?
Are we looking for subarrays that are contiguous? (e.g., must the elements be adjacent in the original array)
If the sum of imbalance numbers exceeds the maximum representable value for a standard integer type, how should I handle potential overflow?

Brute Force Solution

Approach

The goal is to find the total "imbalance" across all possible groups of numbers we can make from the original list. The brute force way to do this means we will look at every single possible group of numbers, one by one.

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

First, consider the single number at the very beginning of the list. Calculate the imbalance of this single-number group. An imbalance is defined as the absolute difference between any two numbers in the subarray that is greater than 1.
Next, consider the first two numbers in the list as a group. Calculate the imbalance of this group.
Then, consider the first three numbers, and so on, until you've considered a group that includes all the numbers in the list. Calculate the imbalance for each of these.
Now, start again, but this time begin with the second number in the list. Repeat the process of forming groups of increasing size from this starting point, calculating the imbalance for each group.
Continue this process, each time starting from the next number in the list and considering all the groups that can be formed from that point onward, calculating the imbalance of each group.
Finally, add up all the individual imbalance values that you calculated for each group. This total is the final answer.

Code Implementation

def sum_of_imbalance_numbers_of_all_subarrays(numbers):
    total_imbalance = 0
    array_length = len(numbers)

    for start_index in range(array_length):
        for end_index in range(start_index, array_length):
            sub_array = numbers[start_index : end_index+1]
            
            # Calculate imbalance for each sub-array.
            if len(sub_array) > 1:
                max_value = max(sub_array)
                min_value = min(sub_array)

                # Imbalance is only added when the difference > 1
                if (max_value - min_value) > 1:

                    total_imbalance += 1

    return total_imbalance

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible subarrays. There are O(n^2) subarrays since we choose a start and end index, both ranging from 0 to n-1. For each subarray, we calculate the imbalance which requires comparing each pair of elements within that subarray. In the worst-case, calculating the imbalance for a subarray of size k requires O(k^2) operations, which is O(n^2) in the worst case since k <= n. Therefore, the total time complexity is O(n^2 * n) which simplifies to O(n^3).

Space Complexity

O(1) – The provided solution calculates the imbalance of each subarray on the fly without storing any intermediate subarrays or large data structures. It iterates through all possible subarrays using nested loops, and within each subarray, it computes the imbalance. The space used is limited to a few variables to keep track of the start and end indices of the subarrays, and a variable to accumulate the imbalance sum, all of which are independent of the input size N. Therefore, the space complexity is constant.

Optimal Solution

Approach

Calculating the imbalance number of every subarray and summing them up directly would take too long. Instead, we focus on each number in the original sequence and determine its impact on the overall imbalance count by considering how many subarrays it affects.

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

Consider each number in the sequence, one at a time.
For each number, find all possible ranges of numbers to its left and right that create subarrays where this specific number is either the biggest or smallest within that subarray.
If the number is the biggest, consider how many subarrays contain it where it's the biggest and calculate the imbalance caused by it being the biggest.
If the number is the smallest, consider how many subarrays contain it where it's the smallest and calculate the imbalance caused by it being the smallest.
Subtract those situations from the total possible imbalances within subarrays that number is part of.
Sum up these calculated effects for every number in the sequence to get the total sum of imbalance numbers of all subarrays.

Code Implementation

def sum_imbalance_numbers(numbers):
    total_imbalance = 0
    array_length = len(numbers)

    for i in range(array_length):
        # Iterate through each number to evaluate its effect on imbalance.
        current_number = numbers[i]
        left_count_less = 0
        left_count_greater = 0
        right_count_less = 0
        right_count_greater = 0

        for j in range(i - 1, -1, -1):
            if numbers[j] < current_number:
                left_count_less += 1
            if numbers[j] > current_number:
                left_count_greater += 1

        for j in range(i + 1, array_length):
            if numbers[j] < current_number:
                right_count_less += 1
            if numbers[j] > current_number:
                right_count_greater += 1

        # Calculate contributions to total imbalance based on smaller/larger.
        total_imbalance += (left_count_greater + right_count_greater) - (left_count_less + right_count_less)

    return total_imbalance

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the input array. For each element, it potentially expands outwards to the left and right to determine the ranges where that element is the maximum or minimum. Finding these ranges in the worst case might require examining all other elements to its left and right for each number. This process resembles a nested loop structure, where for each of the n elements, we perform up to n operations to find the boundaries of its influence. Therefore, the time complexity is proportional to n * n, which simplifies to O(n²).

Space Complexity

O(1) – The provided explanation focuses on iterating through the input sequence and performing calculations based on each element's relationships within its subarrays. It does not mention the creation of any auxiliary data structures like lists, maps, or stacks to store intermediate results or track visited elements. The calculations are performed in place or using a few constant-size variables to keep track of ranges and imbalances. Therefore, the space complexity remains constant, independent of the input size N.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as the sum of imbalance numbers for an empty array is defined as 0.
Input array with a single element	Return 0 because a subarray with only one element has an imbalance number of 0.
Input array with two identical elements	The imbalance number of the array is 0 since the absolute difference is 0.
Input array with elements in descending order	This maximizes the imbalance for each subarray, testing the algorithm's ability to correctly calculate differences.
Input array with a large range of integer values (Integer.MAX_VALUE and Integer.MIN_VALUE)	Ensure the difference calculation doesn't cause integer overflow by using long or appropriate data type.
Input array with all identical values	The sum of imbalance numbers should be 0, as all subarrays will have an imbalance of 0.
Large input array (n > 10000) to check for time complexity	Ensure the chosen algorithm has a time complexity of O(n^2) or better to avoid timeouts.
Input array with negative numbers	The algorithm must handle negative numbers correctly when calculating the absolute difference between elements.