---
title: Binary Search
description: 'Half of a sorted range is removed at each step until the target is found.'
hidden: true
---

Binary search is an efficient array search algorithm. It works by narrowing down the search range by half each time. If you have looked up a word in a physical dictionary, you've already used binary search in real life. Let's look at a simple example:

*Given a sorted array of integers and an integer called target, find the element that equals the target and return its index. If the element is not found, return -1.*

The key observation here is that the array is sorted. We pick a random element in the array and compare it to the target.

- If we happen to pick the element that equals the target (how lucky!), then bingo. We don't need to do any more work; we return its index.
- If the element is smaller than the target, then we know the target cannot be found in the section to the left of the current element, since everything to the left is even smaller. So, we discard the current element and everything on the left from the search range.
- If the element is larger than the target, then we know the target cannot be found in the section to the right of the current element, since everything to the right is even larger. So, we discard the current element and everything on the right from the search range.

We repeat this process until we find the target. Instead of picking a random element, we always pick the **middle** element in the current search range. This way, we can discard half of the options and shrink the search range by half each time. This gives us `O(log(N))` runtime.

This idea can be implemented both iteratively and recursively. However, the major difference is that the iterative version of binary search uses `O(1)` memory while the recursive version uses `O(log(N))` memory.

## Implementation

The search range is represented by the `left` and `right` indices that start from both ends of the array and move towards each other as we search. When moving the index, we discard elements and shrink the search range.

Time Complexity: `O(log(n))`


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```python
def binary_search(arr: list[int], target: int) -> int:
    left, right = 0, len(arr) - 1
    # <= because left and right could point to the same element, < would miss it
    while left <= right:
        # double slash for integer division in python 3,
        # we don't have to worry about integer `left + right` overflow
        # since python integers can be arbitrarily large
        mid = (left + right) // 2
        # found target, return its index
        if arr[mid] == target:
            return mid
        # middle less than target, discard left half by making left search boundary `mid + 1`
        if arr[mid] < target:
            left = mid + 1
        # middle greater than target, discard right half by making right search boundary `mid - 1`
        else:
            right = mid - 1
    return -1  # if we get here we didn't hit above return so we didn't find target
```

### Calculating `mid`

Note that when calculating `mid`, if the number of elements is even, there are two elements in the middle. We usually follow the convention of picking the first one, equivalent to doing integer division `(left + right) / 2`.

In most programming languages, we calculate `mid` with `left + floor((right-left) / 2)` to avoid potential integer overflow. However, in Python, we do not need to worry about integer overflow with `left + right` because Python3 integers can be arbitrarily large.

## Deduction

It's essential to understand and deduce the algorithm yourself instead of memorizing it. In an actual interview, interviewers may ask you additional questions to test your understanding, so simply memorizing the algorithm may not be enough to convince the interviewer.

Key elements in writing a correct binary search:

### 1. When to terminate the loop

Make sure the `while` loop includes an equality comparison; otherwise, we might skip the loop and miss the potential match for the edge case of a one-element array.

### 2. Whether/how to update `left` and `right` boundary in the `if` conditions

Consider which side to discard. If `arr[mid]` is smaller than the `target`, we should discard everything on the left by making `left = mid + 1`.

### 3. Should I discard the current element?

For vanilla binary search, we can discard it since it cannot be the final answer if it is not equal to the target. There might be situations where you would want to think twice before discarding the current element. We'll discuss this in the next module.

## When to use binary search

Interestingly, binary search works beyond sorted arrays. **You can use binary search whenever you make a binary decision to shrink the search range.**