Best approach for beginners using Java
IntuitionThe problem requires counting all subarrays where adjacent elements alternate in value. A straightforward way to approach this is to traverse the array
# Intuition
The problem requires counting all subarrays where adjacent elements alternate in value. A straightforward way to approach this is to traverse the array and track contiguous alternating sequences.

# Approach
- Initialize `count` with `n` since each element itself is an alternating subarray.
- Use a variable `length` to track the length of the current alternating sequence.
- Traverse the array from the second element:
  - If the current element differs from the previous one, extend the sequence.
  - Otherwise, reset `length` to 1.
  - Accumulate `length - 1` to `count` at each step.

# Complexity
- Time complexity: $$O(n)$$ 
- Space complexity: $$O(1)$$ 


# Code
```java []
class Solution {
    public long countAlternatingSubarrays(int[] nums) {
        int n = nums.length;
        long count = n;
        int length = 1;
        
        for (int i = 1; i < n; i++) {
            if (nums[i] != nums[i - 1]) {
                length++;
            } else {
                length = 1;
            }
            count += length - 1;
        }
        
        return count;
    }
}

```
['Java']