2653. Sliding Subarray Beauty

Problem Description

The problem presents a scenario where we are given an array nums consisting of n integers, and we are tasked with determining the "beauty" of every contiguous subarray that is of size k. Beauty in this context is defined as the value of the xth smallest integer in the subarray, but with a twist. If the xth smallest number is negative, that is the beauty. If not, or there are fewer than x negative integers in the subarray, then the beauty is 0. The goal is to create an array that contains the beauty of each subarray starting from the beginning of the nums array, such that we end up with n - k + 1 beauties.

It is important to understand that a subarray is considered a continuous part of the original array and cannot be empty.

In simpler terms, this problem is asking us to look at every possible continuous chunk of k elements in the array, count negative numbers within it and if there are at least x negative numbers, find the xth smallest one, otherwise record a beauty of 0.

Intuition

To solve this problem, a natural approach is to utilize a data structure that can keep numbers sorted as they are added or removed. This way, we can maintain the order of elements in each k-sized subarray in real time without having to sort the subarray on every iteration. This is crucial because sorting every subarray would lead to a high computational complexity, making it inefficient for large arrays.

The solution uses the SortedList data structure from the Python sortedcontainers module which helps maintain a sorted list efficiently. Here's how this works:

1. We initialize SortedList with the first k elements of the nums array. This automatically keeps these elements sorted.
2. We record the beauty of the initial subarray. The xth smallest element can be retrieved directly by index x - 1 due to zero-based indexing. If this element is negative, it is the beauty; otherwise, we record 0.
3. We then slide the window of size k across the array from start to finish. In each step:
   • We remove the leftmost (oldest) element of the subarray from the sorted list since it is no longer in the current window.
   • We add the new element (the one that comes into the window from the right) to the sorted list, keeping the list sorted.
   • We then compute and record the beauty of the new subarray in an analogous manner to step 2.
4. Once we have slid the window across the entire array, we will have computed the beauty of each subarray, and we return the collection of beauties recorded.

Using this method, we continuously maintain a sorted window of the current subarray, allowing us to efficiently answer queries about its xth smallest element and compute the beauty without re-sorting the entire subarray each time.

Learn more about Sliding Window patterns.

Solution Approach

The implementation of the solution uses the SortedList data structure that provides the capability to manage a sorted sequence of numbers efficiently. In Python, SortedList handles changes to the sequence, such as adding or removing elements, while maintaining the sorted order. This capability is key to the solution because it allows us to efficiently manage the min-heap property that makes retrieval of the xth smallest number possible in constant time.

Let's walk through a detailed implementation strategy referencing the code provided:

1. We instantiate a SortedList with the first k elements of the nums array. This prepares our first subarray and maintains its elements in sorted order.

   1sl = SortedList(nums[:k])

2. We calculate the beauty of the first subarray immediately. If the (x - 1) index element is negative, this value is recorded as the beauty; otherwise, 0 is recorded. This becomes the first element of our answer list ans.

   1ans = [sl[x - 1] if sl[x - 1] < 0 else 0]

3. Next, we enter a loop that iterates over nums from the kth index to the end:

   1for i in range(k, len(nums)):

   • For each iteration, the leftmost (oldest) element of the previous subarray is removed from SortedList:

   1sl.remove(nums[i - k])

   • Simultaneously, the next element in the array (nums[i]) is added to SortedList, again maintaining sorting:

   1sl.add(nums[i])

   • We then evaluate the beauty of the current subarray: if the (x - 1) index element is negative, this value is recorded as the beauty; otherwise, 0 is recorded:

   1ans.append(sl[x - 1] if sl[x - 1] < 0 else 0)

4. Finally, after the loop completes, ans contains the beauty of each subarray. This result is then returned.

   1return ans

By using the SortedList, we avoid the need to sort each subarray individually, as it takes care of maintaining the sorted order upon insertions and deletions. This results in an efficient solution that can handle the sliding window computation in a way that is much faster than a naive solution that would sort every single subarray.

Example Walkthrough

Let's use a small example to illustrate the solution approach described above. Suppose we are given an array nums with elements [-1, 2, -3, 4, -5] and we need to find the beauty of every subarray of size k = 3, where the beauty is defined as the x = 2nd smallest negative integer in the subarray.

1. Initialize a SortedList with the first k elements of the nums array:

   1sl = SortedList([-1, 2, -3])

   The sorted list would be [-3, -1, 2].

2. Calculate the beauty of the first subarray:

   1ans = [-1 if sl[1] < 0 else 0]

   Since the element at index 1 (remembering zero-based indexing) is -1, which is negative, we record -1 in ans.

   Now, ans = [-1].

3. Slide the window and update the SortedList:

   For the next subarray [2, -3, 4], remove the element -1 from sl and add 4:

   1sl.remove(-1)
   2sl.add(4)

   Now sl becomes [-3, 2, 4]. The element at index 1 is now 2, so the beauty of this subarray is 0.

   Update ans to include the new beauty: ans = [-1, 0].

4. Move the window to the next subarray [4, -5]:

   Remove the leftmost element 2 from sl and add -5:

   1sl.remove(2)
   2sl.add(-5)

   The sorted list now is [-5, -3, 4]. The element at index 1 is -3, which is negative, so this subarray's beauty is -3.

   Update ans to include this new beauty: ans = [-1, 0, -3].

After following these steps, we obtain the final result ans = [-1, 0, -3], which contains the beauty of each contiguous subarray of size k = 3. This array tells us that the first subarray has a beauty value of -1, the second has 0 since there aren't enough negative numbers to consider, and the third subarray has a beauty value of -3.

Using the SortedList, we efficiently calculated the beauty of all subarrays without having to sort each one individually.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by several operations: initializing the SortedList, adding elements to it, removing elements from it, and looking up an element at a specific index within it.

1. Initializing SortedList with nums[:k] takes O(k log(k)) time, as it needs to sort k elements.
2. The for loop runs (n - k) times, where n is the length of nums.
3. Inside the loop, sl.remove(nums[i - k]) takes O(log(k)) time as SortedList maintains the elements in sorted order, so removal is a logarithmic operation.
4. sl.add(nums[i]) also takes O(log(k)) time for the same reason – to maintain the sorted property of the list.
5. Accessing the x-1th smallest item with sl[x - 1] is an O(1) operation because SortedList allows for fast random access.

Combining these operations, the total time complexity for the loop is (n - k) * (2 * log(k)). So the overall time complexity of the code can be expressed as:

O(k log(k) + (n - k) * log(k)) = O(n log(k))

since k log(k) is insignificant compared to (n - k) * log(k) for large n.

Space Complexity

The space complexity is determined by the extra space used for the SortedList and the ans list.

1. The SortedList at any time contains exactly k elements, giving a space complexity of O(k).
2. The ans list will contain n - k + 1 elements (as we insert a new number for every element from nums[k] to nums[n-1]), so this also gives a space complexity of O(n).

Combining these, the dominant space complexity remains O(n) (as we typically consider the worst-case space usage to determine space complexity).

Therefore, the space complexity of the code is O(n).

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