Here's what I did, no need for a second function or an additional for loop. Not sure why the solution recommended an O(N^3) solution when we could instead use two pointers to keep time complexity to O(N^2).

  def searchTriplets(self, arr, target):
    arr.sort()
    count = 0

    for i in range(0, len(arr) - 1):
      j = i + 1         # left pointer
      k = len(arr) - 1  # right pointer

      while j < k:
        if arr[i] + arr[j] + arr[k] < target:
          count += 1
      
        if j == k - 1:
          k = len(arr) - 1
          j += 1
        else:
          k -= 1

    return count