How do you solve Count Subarrays of Length Three With a Condition on LeetCode?

Count Subarrays of Length Three With a Condition (LeetCode #3392) 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: The condition can be simplified to a single check for each valid middle element.

What is the time complexity of Count Subarrays of Length Three With a Condition?

The optimal solution for Count Subarrays of Length Three With a Condition runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only loop through the array once. The space complexity is O(1) since we are using a constant amount of extra space.

What is the brute force solution for Count Subarrays of Length Three With a Condition?

The brute force approach for Count Subarrays of Length Three With a Condition is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible subarray of length 3 to see if it meets the condition. This is straightforward but can be slow for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Subarrays of Length Three With a Condition?

Count Subarrays of Length Three With a Condition uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Count Subarrays of Length Three With a Condition?

Always clarify the problem statement and constraints with the interviewer. Think aloud while coding to show your thought process. Test your solution with edge cases before finalizing.

What are common mistakes when solving Count Subarrays of Length Three With a Condition?

Not considering the bounds of the array when accessing elements. Confusing the condition and making logical errors in checks.

#3392

Count Subarrays of Length Three With a Condition

Easy

ArrayArrayTwo Pointers

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 approach is still straightforward but leverages the fact that we can check the condition in a single pass through the array, reducing the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Loop through the array from index 1 to nums.length - 2 (to ensure we can access nums[i - 1] and nums[i + 1]).
3Step 3: For each index i, check if nums[i - 1] + nums[i + 1] equals 0.5 * nums[i].
4Step 4: If the condition is met, increment the counter.
5Step 5: Return the counter after checking all relevant indices.

solution.py6 lines

1def countSubarrays(nums):
2    count = 0
3    for i in range(1, len(nums) - 1):
4        if nums[i - 1] + nums[i + 1] == 0.5 * nums[i]:
5            count += 1
6    return count

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array once. The space complexity is O(1) since we are using a constant amount of extra space.

1The condition can be simplified to a single check for each valid middle element.
2Understanding the structure of the problem helps in optimizing the solution.

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