### LeetCode Challenge #34. Find First and Last Position of Element in Sorted Array

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 <= 105`
• `-109 <= nums[i] <= 109`
• `nums` is a non-decreasing array.
• `-109 <= target <= 109`
##### Video Solution
###### Java Solution
```				```
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] ans = {-1, -1};
if(nums.length==0){
return ans ;
}
ans[0] = firstOccurence(nums,target);
ans[1] = lastOccurence(nums,target);
return ans ;
}

public int firstOccurence(int[] nums , int target){
int start = 0 ;
int end = nums.length-1;
int ans = -1 ;

while(start<=end){
int mid = (start + end )/2;
if(nums[mid]==target){
ans = mid ;
end = mid-1;
}else if ( target > nums[mid]){
start = mid+1 ;
}else{
end = mid-1 ;
}
}
return ans ;
}

public int lastOccurence(int[] nums , int target){
int start = 0 ;
int end = nums.length-1;
int ans = -1 ;

while(start<=end){
int mid = (start + end )/2;
if(nums[mid]==target){
ans = mid ;
start = mid+1 ;
}else if ( target > nums[mid]){
start = mid+1 ;
}else{
end = mid-1 ;
}
}
return ans ;
}
}
```
```