Problem Description

The problem provides us with an integer array named arr, and our goal is to find the length of the longest turbulent subarray. A turbulent subarray is defined uniquely by its characteristic that the inequality signs between consecutive elements change or "flip" from one pair to the next. Concretely, a subarray is turbulent if for any two adjacent elements within it, one is greater than the next and the next after that is less than its predecessor, or vice versa; this pattern alternates throughout the subarray.

To put it another way, if the indices of the subarray are from i to j, then for the index k within i and j-1, two conditions are defined:

1. When k is odd, arr[k] should be greater than arr[k + 1], and when k is even, arr[k] should be less than arr[k + 1].
2. Or alternatively, when k is odd, arr[k] should be less than arr[k + 1], and when k is even, arr[k] should be greater than arr[k + 1].

Intuition

When approaching this problem, we consider that a turbulent subarray can start at any index and the conditions for a sequence of elements to be turbulent depend solely on the relationship between adjacent elements. Thus, one way to solve this is by iterating through the array and tracking the length of the current turbulent subarray as we go.

The intuition behind the solution lies in a sliding window or two-pointer approach, where we track the current and maximum lengths of valid subarrays. However, we can optimize this strategy by keeping track of the last two comparisons (whether the last was '>' or '<') which can be done with counters that reset when the pattern is broken.

In the provided solution, two variables f and g are used to keep track of the length of two types of turbulent subarrays, one where the first comparison is '<' and the other where the first comparison is '>'. If the current pair of elements continue the turbulence condition, we increment the appropriate counter (f or g), otherwise, we reset it to 1 (the current pair itself can be the beginning of a new subarray). ans is used to keep track of the maximum size found so far.

We iterate through the array using the pairwise utility, which gives us adjacent pairs of elements for easy comparison. For each pair, using the comparison result, we update f and g, then update ans to hold the maximum of ans, f, and g. At the end of the iteration, ans contains the length of the largest turbulent subarray.

The solution leverages a dynamic and continuous update of variables to reflect the current state of the subarray being checked, thus arriving at the final answer without needing to store any intermediate subarrays or additional complex data structures.

Learn more about Dynamic Programming and Sliding Window patterns.

Solution Approach

The implementation of the solution through the provided Python code uses a straightforward approach, mainly utilizing iteration and simple condition checks without the need for additional data structures beyond simple variables. Here’s how it works:

Variables:

• ans: To keep track of the maximum subarray size found during iteration.
• f: To track the length of turbulent subarrays where the first condition (arr[k] < arr[k + 1] when k is even, and arr[k] > arr[k + 1] when k is odd) holds true.
• g: To track the length of turbulent subarrays where the second condition (arr[k] > arr[k + 1] when k is even, and arr[k] < arr[k + 1] when k is odd) holds true.

Iteration:

• The code uses the pairwise function to create a windowed pair of array elements (a, b) for each adjacent pair in arr.
• On each iteration step, we compare the elements a and b in the tuple to determine the continuation or the end of a turbulent subarray.

Condition checks and Updates:

• ff and gg are temporary variables used to calculate the potential new lengths of f and g.
• If a < b and we are considering the odd index for k, f should be incremented, since the condition arr[k] > arr[k + 1] must hold true for k being even; otherwise, f resets to 1 (subarray starts anew).
• Conversely, if a > b for an even index, g should be incremented, as the condition arr[k] < arr[k + 1] must hold true for an odd k; otherwise, g resets to 1.
• After each comparison, f and g are updated with the values from ff and gg, respectively.
• ans is updated with the maximum value between ans, f, and g.

Return Value:

• After the loop completes, the value of ans denotes the longest turbulent subarray that can be found in arr.

This solution approach relies on dynamically updating counters and checking each pair once. It avoids the need for complex data structures or algorithms. Due to its simplicity and only using iteration and basic arithmetic operations, the overall time complexity is O(n), where n is the number of elements in arr.

Example Walkthrough

Let's illustrate the solution approach with a small example using the array arr = [9, 4, 2, 10, 7, 8, 8, 1, 9].

1. Initialize ans = 1, f = 1, and g = 1 because the minimum length of a turbulent subarray is 1 (a single element).

2. Begin iterating from the start of the array using the pairwise function:

   • First pair (9, 4): since 9 > 4, set ff = 1 and gg = f + 1. Now, g becomes 2 (g = gg), because this pair meets the second condition.
   • Next pair (4, 2): 4 > 2, increment g to 3 (g = g + 1), f resets to 1.
   • Next pair (2, 10): now 2 < 10, so we increment f to g + 1, which is now 4, and reset g to 1.
   • Continue with this pattern for subsequent pairs.

3. At each step, update ans to the maximum of ans, f, and g to ensure it always holds the maximum subarray length found so far.

Following through with the example sequence:

• After processing (10, 7), f is reset to 1 since 10 > 7, g = f + 1 = 2, and ans is updated to max(ans, f, g) = 4.
• After (7, 8), f = g + 1 = 3, g is reset to 1, update ans.
• After (8, 8), both f and g reset to 1, because the elements are equal and that does not meet any turbulent condition.
• After (8, 1), g = f + 1 = 2, f is reset to 1, update ans.
• After (1, 9), f = g + 1 = 3, g is reset to 1, update ans which remains 4, since f is not greater than the current ans.

4. Finally, the iteration ends and we return ans, which is still 4, representing the longest turbulent subarray [4, 2, 10, 7].

This example showcases the elegance and simplicity of the solution. Without storing any subarrays or using complex structures, the algorithm dynamically keeps track of the state and updates it with each element pair. Overall, this clever approach ensures optimal performance with linear time complexity, only passing through the array once.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is primarily governed by the single for loop that iterates through the pairwise elements of the array arr. In Python, the pairwise utility creates an iterator that aggregates elements from each iterable sequence. Considering an input array of size n, there will be n - 1 iterations (since pairwise generates tuples of two consecutive elements).

Within the for loop, there are constant time operations like comparisons, assignments, and a call to the max function with three arguments. These operations contribute O(1) time complexity for each iteration.

Therefore, the total time complexity of this algorithm is O(n).

Space Complexity

The space complexity of the algorithm is determined by the extra space used besides the input array. In this case, only a constant number of integer variables (ans, f, g, ff, and gg) are used. No auxiliary data structures that scale with the input size are utilized.

Thus, the space complexity is O(1), which means it uses a constant amount of space regardless of the input array size.

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