Problem Description

You are given an array nums containing n integers. Your goal is to find the maximum value of k such that you can find two adjacent subarrays, each of length k, where both subarrays are strictly increasing.

More specifically:

• You need to find two subarrays starting at indices a and b
• These subarrays must be adjacent, meaning b = a + k (the second subarray starts immediately after the first one ends)
• Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] must be strictly increasing (each element is greater than the previous one)
• You want to find the maximum possible value of k that satisfies these conditions

For example, if nums = [1, 2, 3, 2, 3, 4]:

• You can find two adjacent strictly increasing subarrays of length 2: [1, 2] and [3, 2] - but wait, [3, 2] is not strictly increasing
• You can find [2, 3] starting at index 1 and [2, 3, 4] starting at index 3, but they're not adjacent
• The maximum k would be 1, as you can find adjacent subarrays [3] and [2] (both of length 1 are trivially strictly increasing)

The solution approach tracks the lengths of consecutive strictly increasing sequences in the array. For each position, it maintains:

• cur: the current length of the strictly increasing sequence
• pre: the length of the previous strictly increasing sequence
• ans: the maximum k found so far

When a strictly increasing sequence ends (either at the array's end or when nums[i] >= nums[i+1]), the code updates the answer by considering:

• cur // 2: the possibility of splitting the current increasing sequence into two equal adjacent parts
• min(pre, cur): the possibility of using the end of the previous sequence and the beginning of the current sequence as the two adjacent subarrays

Intuition

The key insight is that two adjacent strictly increasing subarrays of length k can only come from two scenarios:

1. Within a single increasing sequence: If we have a long strictly increasing sequence, we can split it into two adjacent parts. For a sequence of length n, the maximum k we can achieve is n // 2 (we split it as evenly as possible).

2. Across two consecutive increasing sequences: When one increasing sequence ends and another begins, we can take the tail of the first sequence and the head of the second sequence as our two adjacent subarrays. The maximum k here is limited by the shorter of the two sequences, hence min(pre, cur).

To understand why this works, consider that as we traverse the array, we're essentially identifying all strictly increasing segments. For example, in [1, 2, 3, 1, 2, 3, 4], we have two segments: [1, 2, 3] and [1, 2, 3, 4].

At each boundary between segments (when nums[i] >= nums[i+1]), we have an opportunity to check:

• Can we split the current segment into two equal adjacent parts? This gives us cur // 2
• Can we use parts from both the previous and current segments? This gives us min(pre, cur)

By tracking these values throughout our traversal and keeping the maximum, we ensure we don't miss any valid configuration. The algorithm is efficient because it processes each element exactly once, accumulating information about increasing sequences as it goes.

The beauty of this approach is that it reduces a seemingly complex problem about finding specific subarray patterns into a simpler problem of tracking the lengths of increasing sequences and making local decisions at their boundaries.

Learn more about Binary Search patterns.

Solution Approach

The solution uses a single-pass algorithm with three variables to track the state as we iterate through the array:

Variables:

• ans: Stores the maximum value of k found so far
• pre: Length of the previous strictly increasing sequence
• cur: Length of the current strictly increasing sequence being built

Algorithm Steps:

1. Initialize all variables to 0: ans = pre = cur = 0

2. Iterate through the array using enumeration to get both index i and value x:

   • Increment cur by 1 for each element (extending the current increasing sequence)

3. Check for sequence boundary at each position:

   • A boundary occurs when either:
     • We reach the last element (i == len(nums) - 1), or
     • The current element is greater than or equal to the next element (x >= nums[i + 1])

4. When a boundary is found, update the answer:

   • Calculate max(ans, cur // 2, min(pre, cur))
     • cur // 2: Maximum k if we split the current sequence into two adjacent parts
     • min(pre, cur): Maximum k using the tail of previous and head of current sequence
     • Keep the maximum among these and the existing ans

5. Update state for next iteration:

   • Set pre = cur (current sequence becomes the previous)
   • Reset cur = 0 (start counting the next sequence)

Example Walkthrough:

For nums = [2, 5, 7, 8, 9, 2, 3, 4, 3, 1]:

• Sequence 1: [2, 5, 7, 8, 9] (length 5)
  • At boundary, ans = max(0, 5//2, min(0, 5)) = 2
  • Update: pre = 5, cur = 0
• Sequence 2: [2, 3, 4] (length 3)
  • At boundary, ans = max(2, 3//2, min(5, 3)) = 3
  • Update: pre = 3, cur = 0
• Sequence 3: [3] and [1] (each length 1)
  • Final ans = 3

The algorithm efficiently finds the maximum k in O(n) time with O(1) space complexity.

Example Walkthrough

Let's walk through the algorithm with nums = [1, 3, 5, 2, 4, 6, 7, 8].

Initial state: ans = 0, pre = 0, cur = 0

Step 1: Process index 0, value 1

• cur = 1 (start counting first sequence)
• Check boundary: 1 < 3 (next element), no boundary
• Continue

Step 2: Process index 1, value 3

• cur = 2 (sequence is [1, 3])
• Check boundary: 3 < 5, no boundary
• Continue

Step 3: Process index 2, value 5

• cur = 3 (sequence is [1, 3, 5])
• Check boundary: 5 > 2 (next element), boundary found!
• Update answer: ans = max(0, 3//2, min(0, 3)) = max(0, 1, 0) = 1
• Update state: pre = 3, cur = 0

Step 4: Process index 3, value 2

• cur = 1 (start new sequence [2])
• Check boundary: 2 < 4, no boundary
• Continue

Step 5: Process index 4, value 4

• cur = 2 (sequence is [2, 4])
• Check boundary: 4 < 6, no boundary
• Continue

Step 6: Process index 5, value 6

• cur = 3 (sequence is [2, 4, 6])
• Check boundary: 6 < 7, no boundary
• Continue

Step 7: Process index 6, value 7

• cur = 4 (sequence is [2, 4, 6, 7])
• Check boundary: 7 < 8, no boundary
• Continue

Step 8: Process index 7, value 8

• cur = 5 (sequence is [2, 4, 6, 7, 8])
• Check boundary: Last element, boundary found!
• Update answer: ans = max(1, 5//2, min(3, 5)) = max(1, 2, 3) = 3

Result: Maximum k = 3

This corresponds to using the tail of the first sequence [1, 3, 5] and the head of the second sequence [2, 4, 6] as our two adjacent subarrays of length 3.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input array nums.

The algorithm performs a single pass through the array using one for loop. For each element, it performs constant time operations:

• Incrementing cur by 1
• Checking if current index is the last element or comparing two adjacent elements
• Computing the maximum of three values: ans, cur // 2, and min(pre, cur)
• Updating pre and cur variables

Since each element is visited exactly once and all operations within the loop take O(1) time, the overall time complexity is O(n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Three integer variables: ans, pre, and cur
• Loop variables: i and x

These variables don't scale with the input size, so the space complexity is constant O(1).

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

Common Pitfalls

1. Misunderstanding the "Adjacent Subarrays" Requirement

A common mistake is thinking that adjacent subarrays can overlap or have gaps between them. The problem specifically requires that if the first subarray ends at index a + k - 1, the second must start exactly at index a + k.

Incorrect Understanding:

• Thinking subarrays like [1, 2, 3] and [2, 3, 4] are adjacent (they overlap)
• Thinking subarrays like [1, 2] and [4, 5] are adjacent (they have a gap)

Solution: Ensure you understand that b = a + k means the second subarray starts immediately after the first one ends, with no overlap or gap.

2. Off-by-One Error in Boundary Detection

The condition if index == len(nums) - 1 or value >= nums[index + 1] can be error-prone. Developers might write:

• if index == len(nums) or ... (causes IndexError)
• if index > len(nums) - 1 or ... (misses the last element)
• Forget to check the last element entirely

Solution: Always verify boundary conditions carefully. The check should be index == len(nums) - 1 to properly handle the last element.

3. Incorrectly Handling Single-Element Sequences

When encountering non-increasing elements like [5, 3, 7], the algorithm creates single-element "increasing" sequences. Some might think single elements shouldn't count as strictly increasing.

Incorrect Approach:

# Wrong: Trying to skip single elements
if curr_subarray_length == 1:
    continue  # This breaks the algorithm

Solution: Single-element sequences are valid (trivially strictly increasing). The algorithm correctly handles them as they contribute to valid k=1 solutions.

4. Forgetting to Reset Current Length

A critical mistake is forgetting to reset curr_subarray_length = 0 after processing a boundary. This would cause the counter to keep accumulating incorrectly.

Incorrect Code:

if index == len(nums) - 1 or value >= nums[index + 1]:
    max_length = max(max_length, curr_subarray_length // 2, min(prev_subarray_length, curr_subarray_length))
    prev_subarray_length = curr_subarray_length
    # Missing: curr_subarray_length = 0

Solution: Always reset curr_subarray_length = 0 after updating prev_subarray_length to start counting the next sequence fresh.

5. Not Considering Both Cases for Maximum k

Some solutions might only consider splitting a single increasing sequence (cur // 2) or only consider consecutive sequences (min(pre, cur)), missing valid cases.

Incomplete Solution:

# Wrong: Only considering consecutive sequences
max_length = max(max_length, min(prev_subarray_length, curr_subarray_length))

Solution: Always check both possibilities:

• curr_subarray_length // 2: Can we split the current sequence?
• min(prev_subarray_length, curr_subarray_length): Can we use two consecutive sequences?

The maximum of these gives the correct answer.