## Question

Given an array of integers `nums`

sorted in non-decreasing order, find the starting and ending position of a given `target`

value.

If `target`

is not found in the array, return `[-1, -1]`

.

You must write an algorithm with `O(log n)`

runtime complexity.

**Example 1:**

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

**Example 2:**

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

**Example 3:**

Input:nums = [], target = 0Output:[-1,-1]

**Constraints:**

`0 <= nums.length <= 10`

^{5}`-10`

^{9}<= nums[i] <= 10^{9}`nums`

is a non-decreasing array.`-10`

^{9}<= target <= 10^{9}

## Python Solution

class Solution: def searchRange(self, nums: List[int], target: int) -> List[int]: l = len(nums) i = 0 j = l-1 if l==0: return [-1,-1] while(i<j): mid = i + (j-i)//2 if nums[mid]<target: i = mid+1 else: j = mid if nums[i]!=target: return [-1,-1] start=i j=l-1 while(i<j): mid = i + (j-i)//2 + 1 if nums[mid]>target: j = mid-1 else: i = mid end=i return [start,end]