input: from typing import List
  
  def find_alternating_subarrays(nums: List[int]) -> int:
      """
      Given a binary array nums, count the number of subarrays in which no two successive elements are identical.
  
      Return the total count of such alternating subarrays in nums.
  
      Examples:
      >>> nums = [0,1,1,1]
      5
      >>> nums = [1,0,1,0]
      10
  
      Constraints:
      1 <= len(nums) <= 10^5
      Each nums[i] is either 0 or 1.
      """
output:     n = len(nums)
      count = 0
      
      # Iterate over all possible starting points of the subarray
      for i in range(n):
          # Initialize the alternating flag
          is_alternating = True
          # Check the subarray starting at i and ending at j
          for j in range(i, n):
              if j > i and nums[j] == nums[j-1]:
                  is_alternating = False
              if is_alternating:
                  count += 1
              else:
                  break
      
      return count