Find first and Last position of Element in sorted array

#you can reduce the space complexity by reducing the recursion to while loop.
def first_pos(arr, i, j, target):
    if i <= j:
        mid = (i + j) // 2
        if arr[mid] == target and (mid == 0 or arr[mid - 1] != target):
            return mid
        if arr[mid] >= target: #even if target is in mid and mid-1 both position,it will manage
            return first_pos(arr, i, mid - 1, target)
        else:
            return first_pos(arr, mid + 1, j, target)
    return -1

def last_pos(arr, i, j, target):
    if i <= j:
        mid = (i + j) // 2
        if arr[mid] == target and (mid == len(arr) - 1 or arr[mid + 1] != target):
            return mid
        if arr[mid] <= target: #even if target is in mid and mid+1 both position,it will manage
            return last_pos(arr, mid + 1, j, target)
        else:
            return last_pos(arr, i, mid - 1, target)
    return -1

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        a=first_pos(nums,0,len(nums)-1,target)
        b=last_pos(nums,0,len(nums)-1,target)
        return [a,b]