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

Adjacent Increasing Subarrays Detection I (LeetCode #3349) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Adjacent subarrays must be checked carefully to ensure both are strictly increasing.

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

The brute force approach for Adjacent Increasing Subarrays Detection I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible pairs of adjacent subarrays of length k to see if both are strictly increasing. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

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

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

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

Clarify the problem requirements before jumping into coding. Think about edge cases, such as the smallest and largest possible inputs. Explain your thought process while coding to demonstrate your understanding.

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

Not checking the bounds correctly when accessing array elements. Assuming that checking one subarray is enough without considering the adjacent one.

#3349

Adjacent Increasing Subarrays Detection I

Easy

ArrayTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass to check for strictly increasing subarrays while keeping track of the previous subarray's status. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to track if the first subarray of length k is strictly increasing.
2Step 2: Loop through the array to check if the first subarray is increasing.
3Step 3: If the first subarray is increasing, continue to check the next subarray starting at index k.
4Step 4: If both subarrays are increasing, return true.
5Step 5: If no valid pairs are found by the end of the loop, return false.

solution.py15 lines

1def has_adjacent_increasing_subarrays(nums, k):
2    n = len(nums)
3    if n < 2 * k:
4        return False
5    first_increasing = True
6    for i in range(k - 1):
7        if nums[i] >= nums[i + 1]:
8            first_increasing = False
9            break
10    if not first_increasing:
11        return False
12    for i in range(k, 2 * k - 1):
13        if nums[i] >= nums[i + 1]:
14            return False
15    return True

ℹ

Complexity note: The complexity is O(n) because we only make a single pass through the array to check both subarrays, which is much more efficient than the brute force approach.

1Adjacent subarrays must be checked carefully to ensure both are strictly increasing.
2Understanding the properties of increasing sequences can help optimize the solution.

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