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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 = 8
**Output:** [3,4]

**Example 2:**

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

**Example 3:**

**Input:** nums = [], target = 0
**Output:** [-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}

```
public class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = new int[2];
result[0] = findFirst(nums, target);
result[1] = findLast(nums, target);
return result;
}
private int findFirst(int[] nums, int target){
int idx = -1;
int start = 0;
int end = nums.length - 1;
while(start <= end){
int mid = (start + end) / 2;
if(nums[mid] >= target){
end = mid - 1;
}else{
start = mid + 1;
}
if(nums[mid] == target) idx = mid;
}
return idx;
}
private int findLast(int[] nums, int target){
int idx = -1;
int start = 0;
int end = nums.length - 1;
while(start <= end){
int mid = (start + end) / 2;
if(nums[mid] <= target){
start = mid + 1;
}else{
end = mid - 1;
}
if(nums[mid] == target) idx = mid;
}
return idx;
}
```

```
vector<int> searchRange(vector<int>& nums, int target) {
int idx1 = lower_bound(nums, target);
int idx2 = lower_bound(nums, target+1)-1;
if (idx1 < nums.size() && nums[idx1] == target)
return {idx1, idx2};
else
return {-1, -1};
}
int lower_bound(vector<int>& nums, int target) {
int l = 0, r = nums.size()-1;
while (l <= r) {
int mid = (r-l)/2+l;
if (nums[mid] < target)
l = mid+1;
else
r = mid-1;
}
return l;
}
```

```
def searchRange(self, nums, target):
def binarySearchLeft(A, x):
left, right = 0, len(A) - 1
while left <= right:
mid = (left + right) / 2
if x > A[mid]: left = mid + 1
else: right = mid - 1
return left
def binarySearchRight(A, x):
left, right = 0, len(A) - 1
while left <= right:
mid = (left + right) / 2
if x >= A[mid]: left = mid + 1
else: right = mid - 1
return right
left, right = binarySearchLeft(nums, target), binarySearchRight(nums, target)
return (left, right) if left <= right else [-1, -1]
```

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