Problem Description

The problem requires calculating the "imbalance number" of all subarrays of a given array nums. The imbalance number of an array is defined as the count of instances where, in the sorted version of an array, consecutive elements differ by more than 1. Essentially, it measures how many times a pair of neighboring elements in a sorted list have a gap larger than 1.

To compute the sum of imbalance numbers of all subarrays, we must consider every continuous slice of the original array, sort each one, and calculate its imbalance number, then add these up for all possible subarrays.

A simple example to illustrate this concept would be:

Given the array [1,2,4], it has three subarrays: [1], [1,2], [1,2,4]. Their sorted versions are [1], [1,2], and [1,2,4] respectively. There is no pair of neighboring elements that differ by more than 1 in the first two subarrays, so their imbalance numbers are 0. However, in the last subarray, 2 and 4 have a difference of 2, which is greater than 1; thus, its imbalance number is 1. So the sum of all imbalance numbers is 0 + 0 + 1 = 1.

Intuition

To solve this problem, we adopt an approach that iteratively considers every possible subarray starting from each index. We initialize the imbalance number for each subarray as zero and update it as we grow the subarray window.

The solution involves two nested loops. The outer loop fixes the starting point of the subarray, while the inner loop extends the subarray by incrementing the endpoint. This way, we examine all contiguous subarrays of the array nums.

For each element added to the current subarray, we maintain a sorted list of all elements in the subarray to quickly determine the neighbors of the new element. This ordered list is crucial because it allows us to assess imbalances in constant or logarithmic time, rather than sorting the subarray every time which would be computationally expensive.

When a new number is added:

• We find its direct neighbors in the sorted list to see if it introduces any new imbalances.
• If the new number creates an imbalance with either (or both) of its neighbors, we update our imbalance counter.
• If the new number is inserted between two other numbers that were previously causing an imbalance, the presence of the new number may fix this imbalance, and we decrease our counter.

The variable cnt maintains the imbalance number of the current subarray, which is updated on each iteration based on the conditions explained above. After iterating through all subarrays, the ans variable holds the sum of the imbalance numbers from all subarrays, which is returned as the solution.

Thus, this algorithm effectively combines the enumeration of subarrays with the maintenance of a dynamically-updated sorted list, which leads to a much more efficient calculation of the sum of imbalance numbers than a naive approach.

Solution Approach

The implementation capitalizes on an important data structure—an ordered set. In the Python solution provided, the SortedList from the sortedcontainers module serves as a replacement for an ordered set that is not natively available in Python. Using this data structure, the algorithm keeps track of a subarray's elements in sorted order — an essential step for calculating the imbalance number efficiently.

For every subarray processed, the solution steps through the following algorithm:

1. Initialize an empty SortedList called sl and a counter cnt for the imbalance number.
2. Iterate over the array with two pointers, marked by the indices i and j, where i is fixed by the outer loop and j is varied by the inner loop. Each iteration of the inner loop examines a new subarray extending from i to j.
3. Before adding the new element at j to the subarray:
   • Use the bisect_left method to find the position k in the sorted list where the new element would be inserted. This gives us the direct higher neighbor in the sorted subarray.
   • Determine the immediate lower neighbor in the sorted subarray by subtracting one from the index k, resulting in index h.
4. Check for imbalances:
   • If there is an element at index k, and the difference between this element and nums[j] is greater than 1, increment cnt as a new imbalance is introduced.
   • Similarly, if there is an element at index h, and the difference between nums[j] and this element is greater than 1, increment cnt for this new imbalance.
   • If both conditions are met, check if the new element nums[j] is filling a gap between sl[h] and sl[k] that was previously an imbalance; if so, decrement cnt since an imbalance is resolved.
5. Insert the new element nums[j] into the sorted list to update the subarray.
6. Add the current cnt to a running total ans which accumulates the imbalance numbers for all subarrays examined so far.
7. Repeat steps 3-6 for each j until all subarrays starting with index i have been processed.

After the outer loop concludes, ans holds the final result: the sum of imbalance numbers for all subarrays in nums. This algorithm leverages the efficient search and insertion operations provided by SortedList to calculate the imbalance number dynamically as each element is considered, thus avoiding the need for repeated sorting of each subarray.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach described above using the array nums = [3, 1, 4, 2].

We need to find the sum of imbalance numbers for all subarrays of this array.

1. Start with an empty SortedList called sl and a variable cnt initialized to 0. The variable ans will accumulate the sum of imbalance numbers and is also initialized to 0.

2. Begin with the outer loop, where i starts at 0. That is the starting point of our subarray.

3. Now, let's move to the inner loop, where j will go from i to the length of nums - 1. For each value of j, we follow the steps below.

4. At i = 0, j = 0, the subarray is [3]. Add 3 to sl (which was empty). There are no neighbors to check for imbalance as this is the first element. No need to update cnt, and add cnt (which is 0) to ans.

5. At i = 0, j = 1, the subarray is [3, 1]. We plan to insert 1 into sl, but before that, we check for imbalances:

   • The position k found by bisect_left for 1 is 0 as 1 is smaller than 3.
   • There is no element at h = k - 1 since k is 0.
   • Position k points to 3, and since abs(3 - 1) > 1, we increment cnt to 1.
   • Insert 1 into sl, which is now [1, 3].
   • Add the current cnt (which is 1) to ans.

6. At i = 0, j = 2, the subarray is [3, 1, 4]. Inserting 4:

   • The position k for 4 is 2 as 4 should be inserted after 3.
   • The lower neighbor index h is 1, which corresponds to 3 in sl.
   • Position k is out of bounds, so there's no upper neighbor to compare with 4.
   • For the lower neighbor (sl[h] is 3), abs(4 - 3) = 1 which doesn't increment cnt.
   • Insert 4 into sl, now [1, 3, 4].
   • Add the current cnt (still 1) to ans.

7. At i = 0, j = 3, the subarray is [3, 1, 4, 2], insert 2:

   • The position k for 2 is 2 as 2 should be inserted between 1 and 3.
   • The lower neighbor index h is 1, which corresponds to 1 in sl.
   • There is an imbalance between 2 and 3 since abs(3 - 2) = 1. No update to cnt.
   • There is no imbalance between 2 and 1, since abs(2 - 1) = 1. No update to cnt.
   • No existing imbalance is corrected by adding 2 because there were no elements between 1 and 3 causing an imbalance before.
   • Insert 2 into sl, now [1, 2, 3, 4].
   • Add the current cnt (still 1) to ans.

8. Now, the outer loop moves to i = 1 and the process repeats with a new starting point, checking for potential imbalances as elements are added and updating the counters accordingly.

After all iterations of both i and j, ans holds the sum of the imbalance numbers for all subarrays in nums.

This approach dynamically calculates the imbalance number as each element is considered and efficiently updates both the imbalance counters and the sorted list of the subarray elements without needing to sort subarrays repeatedly.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code has a nested loop where the outer loop runs n times, and the inner loop runs up to n times as well, where n is the length of the input array nums. For each iteration of the inner loop, the code executes operations that include binary searches and insertions into a SortedList. Each of these operations (binary search and insertion) in a SortedList has a time complexity of O(log n).

This means that for each i, we are doing n - i insertions, and up to n - i binary searches, each taking O(log n) time. The sum of n - i over all valid i is the sum of the first n integers, which is (n(n+1))/2, simplifying the total number of operations to (n^2 + n)/2. Since each operation is O(log n), we multiply the number of operations by the cost of each operation to get the overall time complexity.

Therefore, the time complexity is O(n^2 * log n) because the dominant factor is the nested loop with the SortedList operations, and the constant factors can be ignored in the Big O notation.

Space Complexity

The space complexity of the code can be analyzed by accounting for the space taken up by the SortedList. This list can grow up to n elements in size, where n is the length of the input array nums. No other data structures in the code use more space than this, and the space used by variables such as i, j, k, h, and cnt is negligible compared to the space taken up by the SortedList.

Therefore, the space complexity is O(n), as the SortedList is the data structure occupying the most space and it grows linearly with the input size.

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