# Quick Access

# Description

## Problem

Given a sorted sequence,find the exact position of a given element.

A simple approach is to do a linear search (check each element one by one).
The time complexity of a **linear search is O(n)**.
Another approach to perform the same task in **O(log n) is using Binary Search**.

Considering a sorted list that contains **1 million items**:

- If we perform a linear search we could have to make **1 million iterations and
1 million comparisons**

- With a binary search we will find it for sure with no more than
**20 iterations and 20 comparisons (asymptotically 5000 times faster!)**

## Applications

- Find data index, more importantly
**within huge sequences database**. - Key idea in
**binary search trees**, a fundamental data structure. - It is used to estimate the
**square roots of numbers**(e.g. Newton's method).

### Homonyms

- Dichotomic search
- Half-interval search
- Logarithmic search
- Binary chop

# How To Build

Binary search is a decrease and conquer search algorithm that can be used on a sorted array.
It operates by determining whether the search value is less or
higher than the middle value.
It then **loops on** the lower or upper **half of the sequence**
until either the value is found or not (half is an empty sequence).

## Steps

- 1.
**Compare**search value with the middle element. - 2. If they
**match**, we are done:**return the middle index**. - 3. If the
**search value is higher**: loop within the**upper half sequence**. - 4. If the
**search value is smaller**: loop within the**lower half sequence**.

## Pseudo-Code

Index BinarySearch(Iterator begin, Iterator end, const T key) { auto index = -1; auto middle = begin + distance(begin, end) / 2; // While there is still objects between the two iterators and no object has been found yet while(begin < end && index < 0) { if (IsEqual(key, middle)) // Object Found: Retrieve index index = position(middle); else if (key > middle) // Search key within upper sequence begin = middle + 1; else // Search key within lower sequence end = middle; middle = begin + distance(begin, end) / 2; } return index; }

# Visualization Analysis

As the analysis visualization is quite trivial: it has been included within the header image carousel of this article.

# Complexity

## Time

### Best

The best configuration occurs when
**the first middle point selected is the search value;**
leading to a complexity of **O( 1)**.

### Average / Worst

The worst case scenario occurs when the **search value is not in the sequence**.

After the first look-up, we loop the search on the lower/upper half and so on.
Let us put this into a recurrence relation:
$$T(n) = O(1) + T(\frac{n}{2})$$
Using the master theorem for divide-and-conquer recurrences: **T(n) = O(log n)**.

## Space

Binary search does not use any buffer nor does make any recursion. Thus, it requires
**O(1)** space in all case.