How do you solve Adjacent Increasing Subarrays Detection II on LeetCode?

Adjacent Increasing Subarrays Detection II (LeetCode #3350) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying strictly increasing subarrays can be done in a single pass.

What is the time complexity of Adjacent Increasing Subarrays Detection II?

The optimal solution for Adjacent Increasing Subarrays Detection II runs in O(n) time and uses O(n) space. This complexity is efficient because we only traverse the array a couple of times, once to find lengths and once to find the maximum k.

What is the brute force solution for Adjacent Increasing Subarrays Detection II?

The brute force approach for Adjacent Increasing Subarrays Detection II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible starting index for two adjacent subarrays of length k. We verify if both subarrays are strictly increasing, incrementing k until we can no longer find such pairs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Adjacent Increasing Subarrays Detection II?

Adjacent Increasing Subarrays Detection II uses the following concepts: Array, Binary Search. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Adjacent Increasing Subarrays Detection II?

Always clarify the problem requirements, especially about adjacency. Discuss your thought process and potential approaches before coding. Test your code with edge cases to ensure robustness.

What are common mistakes when solving Adjacent Increasing Subarrays Detection II?

Not considering edge cases where there are no strictly increasing subarrays. Overlooking the need for adjacent subarrays when calculating k.

#3350

Adjacent Increasing Subarrays Detection II

Medium

ArrayBinary SearchArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach identifies the lengths of all strictly increasing subarrays in a single pass, storing their lengths. We then check for adjacent pairs of these lengths to find the maximum k efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create an array lengths to store the lengths of all strictly increasing subarrays.
2Step 2: Traverse the nums array and populate the lengths array with the lengths of increasing sequences.
3Step 3: Iterate through the lengths array to find the maximum k such that lengths[i] and lengths[i+1] exist and are both greater than or equal to k.

solution.py17 lines

1def maxAdjacentIncreasingSubarrays(nums):
2    n = len(nums)
3    lengths = []
4    length = 1
5    for i in range(1, n):
6        if nums[i] > nums[i - 1]:
7            length += 1
8        else:
9            if length > 1:
10                lengths.append(length)
11            length = 1
12    if length > 1:
13        lengths.append(length)
14    max_k = 0
15    for i in range(len(lengths) - 1):
16        max_k = max(max_k, min(lengths[i], lengths[i + 1]))
17    return max_k

ℹ

Complexity note: This complexity is efficient because we only traverse the array a couple of times, once to find lengths and once to find the maximum k.

1Identifying strictly increasing subarrays can be done in a single pass.
2Adjacent subarrays must be checked for their lengths to find the maximum k.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.