2863. Maximum Length of Semi-Decreasing Subarrays

Problem Description

In this problem, we are given an array nums consisting of integer elements. Our goal is to determine the length of the longest contiguous subarray that meets a specific criterion called "semi-decreasing." A subarray is considered semi-decreasing if it starts with an element strictly greater than the element it ends with. Additionally, we should return 0 if no such subarray exists in nums.

To illustrate, consider the following examples:

• If nums = [5,3,1,2,4], the longest semi-decreasing subarray is [5,3,1], and its length is 3.
• In the case of nums = [1,2,3,4,5], there are no semi-decreasing subarrays, so the answer is 0.

The problem is essentially asking us to find the maximum distance between the starting index of the highest element and the ending index of a lower element, ensuring that this sequence forms a subarray. We must handle this task efficiently, as the naive approach of examining every possible subarray could prove to be too slow.

Intuition

To arrive at the solution, we must track the valuable information that can help us identify semi-decreasing subarrays. Recognizing that a semi-decreasing subarray's first element must be strictly greater than its last element, we can sort the unique elements of the array in reverse (from largest to smallest) and keep track of their indices.

A Python dictionary (here d) can be employed to store the indices of each unique element. By iterating through the array and populating this dictionary, we ensure that for each element, we have a list of all indices where it occurs.

Once this preparation is done, we can iterate over the unique elements in descending order. For each element x, we use the last occurrence of x (since we want to maximize the subarray's length, and a semi-decreasing subarray's last element must be lower than the first) and find the distance to the smallest index k seen so far of elements greater than x. This smallest index k represents the farthest we can go to the left to start a semi-decreasing subarray that ends with the given element x.

The intuition behind maintaining k as the minimum index seen so far is that as we proceed with lower elements, we want to extend our subarray as much to the left as possible. ans represents the length of the longest semi-decreasing subarray found so far, which gets updated each time we find a longer semi-decreasing subarray by subtracting k from the last index of x and adding 1 (to account for array indexing starting from 0).

This approach systematically constructs semi-decreasing subarrays over the original array without the need for exhaustive search, thus providing a much more efficient solution.

Learn more about Sorting patterns.

Solution Approach

The implementation of the solution uses a combination of sorting, hash maps (dictionaries in Python), and linear traversal to find the longest semi-decreasing subarray. Here's a step-by-step breakdown of the algorithm based on the provided solution code:

1. The first step is to iterate through the given array nums and build a hash map (d), where the key is the element value, and the value is a list of indices at which the element occurs. This is achieved with the following line of code:

   1for i, x in enumerate(nums):
   2    d[x].append(i)

   The enumerate function is used to get both the index (i) and the value (x) during iteration.

2. We introduce two variables, ans and k. ans is initialized to 0, representing the length of the longest subarray found so far, and k is initialized to inf (infinity), representing the lowest index seen so far for elements greater than the current one being considered.

3. We sort the keys of the dictionary d in reverse order so we can traverse from the highest element to the lowest. This is important because we are looking for subarrays where the first element is strictly greater than the last element.

   1for x in sorted(d, reverse=True):

4. For each element x (sorted from largest to smallest), we calculate the prospective length of the semi-decreasing subarray ending with x. We achieve this by computing the distance between the last occurrence of x and the smallest index seen so far (k). We add 1 to this distance since the array is zero-indexed:

   1ans = max(ans, d[x][-1] - k + 1)

5. After considering the element x, we update k to hold the smallest index which could potentially be the start of a longer semi-decreasing subarray for the next iterations:

   1k = min(k, d[x][0])

   This means for subsequent smaller elements, we have the information about how far left (to a higher element) we can extend our subarray.

By iterating through the elements in this fashion, we ensure that we are always considering semi-decreasing subarrays (since we move from larger to smaller elements), and we keep extending our reach to the left as much as possible, thus finding the length of the longest such subarray efficiently.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider the following simple array:

1nums = [7, 5, 4, 6, 3]

Applying the algorithm step by step:

Step 1: Build a hash map (d) with element values as keys and a list of their indices as values.

After the first step, d will look like this:

1d = {
2  7: [0],
3  5: [1],
4  4: [2],
5  6: [3],
6  3: [4]
7}

Step 2: Initialize ans to 0 and k to inf.

1ans = 0
2k = inf

Step 3: Sort the keys of dictionary d in reverse order. We get the sorted keys as [7, 6, 5, 4, 3].

Step 4: Iterate over the sorted keys, calculating the potential length of the semi-decreasing subarray and updating ans:

• For key 7, since it's the largest element, there cannot be a semi-decreasing subarray starting with 7. But we use its index 0 to update the smallest index k.

• Update k to min(k, d[7][0]), which will be min(inf, 0) so k becomes 0.

• Now, for key 6, its last index is 3, and the smallest index seen so far is 0. The length of the semi-decreasing subarray would be 3 - 0 + 1 = 4. So ans becomes 4.

• Update k to min(k, d[6][0]), which will be min(0, 3) so k stays 0.

• For key 5, we calculate ans = max(ans, d[5][-1] - k + 1), which is max(4, 1 - 0 + 1) and ans stays 4.

• Update k to min(k, d[5][0]), so k remains 0.

• For key 4, we have ans = max(ans, d[4][-1] - k + 1), which is max(4, 2 - 0 + 1) and ans stays 4.

• Update k to min(k, d[4][0]), so k remains 0.

• Finally, for key 3, we have ans = max(ans, d[3][-1] - k + 1), which is max(4, 4 - 0 + 1), so ans stays 4.

• k does not need to be updated as we have finished the iterations.

Step 5: The largest ans we found was 4, corresponding to the subarray [7, 5, 4, 6], which starts with 7 and ends with 6.

Therefore, the length of the longest semi-decreasing subarray in nums is 4.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by several key operations:

• Enumeration and creation of dictionary: The first for loop goes through all the entries in nums, thus it is O(n) where n is the length of nums.
• Sorting of dictionary keys: The keys of the dictionary are sorted in reverse order. Since the number of unique elements (keys in the dictionary) could be at most n, the sort operation has a complexity of O(u log u) where u is the number of unique keys, with a worst-case scenario being O(n log n).
• Iterating over sorted keys: This iteration potentially goes through all unique keys (at most n), and for each key, the complexity is O(1) since it only accesses the first and last elements of the list of indices associated with each key.

Given these considerations, the overall time complexity is O(n log n) due to the sorting step which is likely the bottleneck.

Space Complexity

The space complexity of the code is influenced by:

• Dictionary storage: The dictionary d stores lists of indices per unique numbers in nums. In the worst case, where all elements are unique, the space complexity due to the dictionary would be O(n).
• Miscellaneous variables: Variables ans, k, and the space for iteration do not depend on size of nums and thus contribute a constant factor.

Hence, the total space complexity is O(n), taking into account the space occupied by the dictionary which is the dominant term.

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