What is the brute force solution for Longest Strictly Increasing or Strictly Decreasing Subarray?

The brute force approach for Longest Strictly Increasing or Strictly Decreasing Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to see if it is strictly increasing or strictly decreasing. 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 Longest Strictly Increasing or Strictly Decreasing Subarray?

Longest Strictly Increasing or Strictly Decreasing Subarray uses the following concepts: Array. The recommended patterns to study are: Two Pointers, Sliding Window, Array.

What are the interview tips for Longest Strictly Increasing or Strictly Decreasing Subarray?

Always think about edge cases and how they affect your logic. Explain your thought process clearly as you code; it helps the interviewer follow your reasoning. Practice dry-running your code with sample inputs to catch errors early.

What are common mistakes when solving Longest Strictly Increasing or Strictly Decreasing Subarray?

Not resetting lengths correctly when encountering equal elements. Failing to consider edge cases, such as arrays with all identical elements.

#3105

Longest Strictly Increasing or Strictly Decreasing Subarray

Easy

ArrayTwo PointersSliding WindowArray

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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)

Instead of checking all subarrays, we can traverse the array once while keeping track of the current length of increasing or decreasing sequences. This reduces the number of checks significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize two variables to track the current length of increasing and decreasing sequences.
2Step 2: Traverse the array and compare each element with the previous one.
3Step 3: If the current element is greater than the previous, increment the increasing length; if less, increment the decreasing length; otherwise, reset both lengths to 1.
4Step 4: Update the maximum length found during the traversal.

solution.py16 lines

1def longest_subarray(nums):
2    max_length = 1
3    inc_length = 1
4    dec_length = 1
5    for i in range(1, len(nums)):
6        if nums[i] > nums[i - 1]:
7            inc_length += 1
8            dec_length = 1
9        elif nums[i] < nums[i - 1]:
10            dec_length += 1
11            inc_length = 1
12        else:
13            inc_length = 1
14            dec_length = 1
15        max_length = max(max_length, inc_length, dec_length)
16    return max_length

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once, making it much more efficient than the brute-force approach. The space complexity is O(1) since we are using a constant amount of extra space.

1Identifying patterns in sequences can help reduce the number of checks needed.
2Tracking lengths of sequences can simplify the problem significantly.

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