# Search for a Range InterviewBit Solution

**Problem: **__Search for a Range__

**Problem Description:**

Given a sorted array of integers **A**(0 based index) of size **N**, find the starting and ending position of a given integer **B** in array **A**.

Your algorithmâ€™s runtime complexity must be in the order of O(log n).

Return an array of size **2**, such that **first element = starting position of B in A** and **second element = ending position of B in A**, if **B** is not found in **A** return **[-1, -1]**.

**Input Format**

`The first argument given is the integer array A. The second argument given is the integer B. `

**Output Format**

` Return an array of size 2, such that first element = starting position of B in A and second element = ending position of B in A, if B is not found in A return [-1, -1]. `

**Constraints**

```
1 <= N <= 10^6
1 <= A[i], B <= 10^9
```

**For Example**

```
Input 1:
A = [5, 7, 7, 8, 8, 10]
B = 8
Output 1:
[3, 4]
Explanation 1:
First occurence of 8 in A is at index 3
Second occurence of 8 in A is at index 4
ans = [3, 4]
Input 2:
A = [5, 17, 100, 111]
B = 3
Output 2:
[-1, -1]
```

## Approach

The solution is to about finding the lower bound and upper bound of the given element.

So we can use two binary search algorithm for each one to find out our desired result.

## Time & Space Complexity

`Time Complexity: O(LogN)`

- Since we have used a binary search algorithm.

`Space Complexity: O(1)`

- Ignoring space taken by input array.

## Solution

Code in C++

## Comments