Grokking Binary Search for Coding Interviews_468 x 60 copy.png

Binary Search (BS) is, arguably, the most popular question in coding interviews. Even when you see that a problem can be solved with BS, it can be tricky to handle all of the indexes and base cases in the solution. What if there was an approach that could work for all BS problems? 
This course was designed from scratch with this one goal in mind. We'll explore one technique that can handle any BS problem. The practice problems in this course were carefully chosen to cover the most frequently asked BS questions in coding interviews. At the end of this course, you will feel confident to handle BS questions during phone or on-site interviews, as well as with all of the follow-ups about its time and space complexity.

Binary Search for Coding Interviews

# Problem statement

Suppose we are given a **sorted** array that only consists of **integers**, where every element appears **twice**, except for one element which appears **once**. Find the element that appears only once.

**Constraints:** 

- $1 \leq nums.length \leq 10^3$
- $0 \leq nums[i] \leq 10^3$

**Example 1:**
```
Input: [1,1,2,3,3,4,4,8,8]    
Output: 2
```
**Example 2:**
```
Input: [3,3,7,7,10,11,11]    
Output: 10
```
**Example 3:**
```
Input: [1,1,2,2,3]    
Output: 3
```
**Example 4:**
```
Input: [1]    
Output: 1
```

# Intuition
 
Since the array contains one element that appears only once, and all of the other elements appear twice, it must always have an odd number of elements.

# Algorithm
    
-   Set the $left$ boundary to $0$, and the right boundary to the size of $nums$.
    
-   While $left + 1 < right$:
    
    -   Take $mid = \frac{(left + right)}{2}$ as a guess. Compare $nums[mid]$ with $nums[mid - 1]$:
        
        -   If $nums[mid]$ != $nums[mid - 1]$ and $(1 + mid$ == size of $nums$ or $nums[mid]$ != $nums[mid + 1])$, the job will be done and we will return $nums[mid]$.
            
        -   Else, if $(mid$ is odd and $nums[mid] == nums[mid - 1])$, or $(mid$ is even, and $nums[mid] == nums[mid + 1])$, we will move the left boundary $left = mid$.
            
        -   Otherwise, we will move the right boundary $right = mid$.
            
-   Return $nums[left]$.

Let's try to use a sorted array with duplicates as the input array.

Difficulty Level: Medium

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Single Element in a Sorted Array

Problem statement