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# Design and Analysis Binary Search

In this chapter, we will discuss another algorithm based on divide and conquer method.

## Problem Statement

Binary search can be performed on a sorted array. In this approach, the index of an element x is determined if the element belongs to the list of elements. If the array is unsorted, linear search is used to determine the position.

## Solution

In this algorithm, we want to find whether element x belongs to a set of numbers stored in an array numbers[]. Where l and r represent the left and right index of a sub-array in which searching operation should be performed.

Algorithm: Binary-Search(numbers[], x, l, r)
if l = r then
return l
else
m := ⌊(l + r) / 2⌋
if x ≤ numbers[m]  then
return Binary-Search(numbers[], x, l, m)
else
return Binary-Search(numbers[], x, m+1, r)


## Analysis

Linear search runs in O(n) time. Whereas binary search produces the result in O(log n) time

Let T(n) be the number of comparisons in worst-case in an array of n elements.

Hence,

$$T(n)=\begin{cases}0 & if\:n= 1\\T(\frac{n}{2})+1 & otherwise\end{cases}$$

Using this recurrence relation $T(n) = log\:n$.

Therefore, binary search uses $O(log\:n)$ time.

## Example

In this example, we are going to search element 63.

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