As others have said, this is a sliding window problem rather than two pointers. Using the sliding window pattern, we get O(N^2) instead of O(N^3).

  def findSubarrays(self, arr, target):
    result = []

    # Use a sliding window
    for start in range(0, len(arr)):
      product = 1.0   # reset for every iteration of start pointer
      subArray = []   # reset new subarray

      # Iterate the end pointer from the start pointer until product exceeds target
      for end in range(start, len(arr)):
        product *= arr[end]
        if product >= target:
          break
         
        subArray.append(arr[end])
        result.append(list(subArray))
    
    return result