How do you solve Count Alternating Subarrays on LeetCode?

Count Alternating Subarrays (LeetCode #3101) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute-force approach involves checking every possible subarray to see if it's alternating. This is straightforward but inefficient, as it requires examining all pairs of start and end indices.

What is the time complexity of Count Alternating Subarrays?

The optimal solution for Count Alternating Subarrays runs in O(n²) time and uses O(1) space. This complexity arises because we are using two nested loops to check every possible subarray, leading to O(n²) comparisons.

What algorithm or data structure is used to solve Count Alternating Subarrays?

Count Alternating Subarrays uses the following concepts: Array, Math. The recommended patterns to study are: Two Pointers, Sliding Window, Dynamic Programming.

What are the interview tips for Count Alternating Subarrays?

Always think about edge cases, such as arrays with all identical elements. Explain your thought process clearly while coding. Consider both time and space complexity when proposing a solution.

What are common mistakes when solving Count Alternating Subarrays?

Forgetting to reset the length when encountering a non-alternating pair. Not considering single-element subarrays.

Count Alternating Subarrays — LeetCode #3101 (Medium)

Tags: Array, Math

Related patterns: Two Pointers, Sliding Window, Dynamic Programming

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute-force approach involves checking every possible subarray to see if it's alternating. This is straightforward but inefficient, as it requires examining all pairs of start and end indices.

This complexity arises because we are using two nested loops to check every possible subarray, leading to O(n²) comparisons.

Step 1: Initialize a variable to count the number of alternating subarrays.
Step 2: Use two nested loops to iterate through all possible subarrays.
Step 3: For each subarray, check if it is alternating by comparing adjacent elements.
Step 4: If it is alternating, increment the count.

For nums = [0, 1, 1, 1]: 1. i = 0, j = 0 -> [0] (count = 1) 2. i = 0, j = 1 -> [0, 1] (count = 2) 3. i = 0, j = 2 -> [0, 1, 1] (not alternating) 4. i = 0, j = 3 -> [0, 1, 1, 1] (not alternating) 5. i = 1, j = 1 -> [1] (count = 3) 6. i = 1, j = 2 -> [1, 1] (not alternating) 7. i = 1, j = 3 -> [1, 1, 1] (not alternating) 8. i = 2, j = 2 -> [1] (count = 4) 9. i = 2, j = 3 -> [1, 1] (not alternating) 10. i = 3, j = 3 -> [1] (count = 5) Final count = 5.

Common Mistakes

Forgetting to reset the length when encountering a non-alternating pair.
Not considering single-element subarrays.

Interview Tips

Always think about edge cases, such as arrays with all identical elements.
Explain your thought process clearly while coding.
Consider both time and space complexity when proposing a solution.

#3101

Count Alternating Subarrays

Medium

Array↗Math↗Two Pointers↗Sliding Window↗Dynamic Programming↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute-force approach involves checking every possible subarray to see if it's alternating. This is straightforward but inefficient, as it requires examining all pairs of start and end indices.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to count the number of alternating subarrays.
2Step 2: Use two nested loops to iterate through all possible subarrays.
3Step 3: For each subarray, check if it is alternating by comparing adjacent elements.
4Step 4: If it is alternating, increment the count.

solution.py13 lines

1# Full working Python code
2
3def countAlternatingSubarrays(nums):
4    count = 0
5    n = len(nums)
6    for i in range(n):
7        for j in range(i, n):
8            if all(nums[k] != nums[k + 1] for k in range(i, j)):
9                count += 1
10    return count
11
12# Example usage
13print(countAlternatingSubarrays([0, 1, 1, 1]))  # Output: 5

ℹ

Complexity note: This complexity arises because we are using two nested loops to check every possible subarray, leading to O(n²) comparisons.

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