Is Count Alternating Subarrays easy or hard?

The problem is rated Medium because the counting insight is not immediately obvious. Many developers initially think about checking every subarray, which leads to O(n^2) time. Recognizing that each alternating streak contributes multiple subarrays enables the optimal O(n) solution.

Count Alternating Subarrays Python/Java solution

In Python or Java, iterate through the array and maintain a variable representing the current alternating streak length. If the current value differs from the previous one, increment the streak; otherwise reset it to 1. Add the streak length to the answer after each step.

How to solve Count Alternating Subarrays in O(n)?

Traverse the array once while maintaining the length of the current alternating run. If nums[i] differs from nums[i-1], increment the run length; otherwise reset it to 1. Add the run length to the total at each step. This counts all subarrays ending at each position in linear time.

What is the best approach for Count Alternating Subarrays?

The best approach is the running count method that tracks the length of the current alternating streak. For each index, if the element differs from the previous one, extend the streak; otherwise reset it. Add the streak length to the answer. This achieves O(n) time and O(1) space.

Is Count Alternating Subarrays asked at Google/Amazon/Meta?

Alternating pattern and subarray counting problems commonly appear in interviews at companies like Amazon, Google, and Meta. Variants test understanding of sliding window patterns, prefix counting, and recognizing how contiguous sequences contribute multiple subarrays.

What data structure is used in Count Alternating Subarrays?

The problem mainly uses array traversal and simple integer counters. No complex data structures are required. The key technique is detecting alternating relationships between adjacent elements while maintaining the current streak length.

What is the time complexity of Count Alternating Subarrays?

The optimal solutions run in O(n) time because the array is scanned once. Each element is compared with its previous neighbor and contributes to the running count of alternating subarrays. Space complexity is O(1) since only a few counters are stored.

Description Hints 3 Solution Video 8:22

3101. Count Alternating Subarrays

Medium57.6% acceptance1 company

You are given a binary array nums.

We call a subarray alternating if no two adjacent elements in the subarray have the same value.

Return the number of alternating subarrays in nums.

Example 1:

Input: nums = [0,1,1,1]

Output: 5

Explanation:

The following subarrays are alternating: [0], [1], [1], [1], and [0,1].

Example 2:

Input: nums = [1,0,1,0]

Output: 10

Explanation:

Every subarray of the array is alternating. There are 10 possible subarrays that we can choose.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is either 0 or 1.

Asked at 1 company

Capital One

Topics

Array Math

Hints open one at a time

Hint 1

Try using dynamic programming.

Hint 2

Let dp[i] be the number of alternating subarrays ending at index i.

Hint 3

The final answer is the sum of dp[i] over all indices i from 0 to n - 1.

8:22

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1class Solution:
2    def countAlternatingSubarrays(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[0,1,1,1]

Input

[1,0,1,0]

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