LeetCode has a Medium coding Problem in Its’ Algorithm Section “Find First And Last Position of Element in Sorted Array”. Today We are going to solve this problem. LeetCode Link of the Problem is HERE

**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.

**Examples**

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

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

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

**Constraints to the problem: **

`0 <= nums.length <= 10`

^{5}`-10`

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

is a non-decreasing array.`-10`

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

**Solution to Find First And Last Position of Element in Sorted Array**

**Skeleton Code** **given by Leetcode on its site.**

```
class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
```

To an ordered array of integers, find start and end positions of the same target, defines the time complexity is O (logn).

Solution: dichotomy, the complexity of the time typical binary search method, first using a binary search for the original array, find out the location of a target, and then find the start and end positions on both sides to search.

**Complete Solution to the given problem**

```
class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
res = [-1, -1]
if nums is None or len(nums) == 0:
return res
size = len(nums)
l, r = 0, size - 1
while l < r:
m = (l + r)/2
if nums[m] < target:
l = m + 1
else:
r = m
if nums[r] != target:
return res
res[0] = r
l, r = 0, size
while l < r:
m = (l + r)/2
if nums[m] <= target:
l = m + 1
else:
r = m
res[1] = l - 1
return res
```

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