find-first-and-last-position-of-element-in-sorted-array

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

My code:
class Solution:
    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        index1,index2=self.find_target(nums,0,len(nums)-1,target)
        return [index1,index2]
    def find_target(self,nums,i,j,target):
        if i>j:
            return [-1,-1]
        if j-i<=1:
            if nums[j]==target and nums[i]==target:
                return [i,j]
            elif nums[j]==target:
                return [j,j]
            elif nums[i]==target:
                return [i,i]
            return [-1,-1]
        medium_index=int((i+j)/2)
        medium=nums[medium_index]
        print('i %d j %d medium %d'%(i,j,medium))
        if medium==target:
            start=medium_index
            while medium==target:
                medium_index+=1
                if medium_index>=len(nums):
                    break
                medium=nums[medium_index]
            medium=nums[start]
            while medium==target:
                print('start %d medium_index %d medium %d'%(start+1,medium_index-1,medium))
                start-=1
                if start<0:
                    break
                medium=nums[start]   
            return [start+1,medium_index-1]
        if medium<target:
            return self.find_target(nums,medium_index,j,target)
        if medium>target:
            return self.find_target(nums,i,medium_index,target)
        return [-1,-1]
        
 Solution:
 https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/