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# 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]
```

```java
public class Solution {
    public int[] searchRange(final int[] A, int B) {
        // Binary searching for 1st occourence
        int left = binarySearch(A, B, false);
        // If we cannot find any occourence
        if (left == -1)
            return new int[] { -1, -1 };
        // Find the last occourence
        return new int[] { left, binarySearch(A, B, true) };
    }

    private int binarySearch(final int[] A, int target, boolean searchLast) {
        int low = 0, high = A.length - 1;
        int result = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (A[mid] == target) {
                result = mid;
                if (searchLast)
                    low = mid + 1;
                else
                    high = mid - 1;
            } else if (target > A[mid])
                low = mid + 1;
            else
                high = mid - 1;
        }
        return result;
    }
}
```