Lecture_Notes/articles/Akash Articles/md/Exponential_Search.md

## Exponential Search

Exponential search is an algorithm for searching in a sorted list. For a sorted list of size $n$, binary search works in $\log{(n)}$, but can we do better than this?

Apparently, no. But what if we use some mechanism to determine a smaller index-range, in which the element we are searching(say $X$) belongs?

For example, we are searching for $8$ in the list ${2,4,5,8,9,10}$. Now, if we use binary search then the searching index-range will be $1$ to $6$. But we can see that if we binary search in the index-range $3$ to $6$ then also we will be able to find $8$.

Note that the number of comparisons for smaller index-range is lesser than the ordinary binary search, this is the main idea behind **Exponential search**.

But how do we determine such an index-range for a given element?

As the name of the algorithm suggests, we are going to use exponentiation. We find an index of the form $2^k$(starting from $k$=$0$), such that the element at that index is greater than $X$.

After that, we do binary search over the index-range $2^{k-1}$ to $2^k$. It is easy to see that $X$ is in the index-range $2^{k-1}$ to $2^k$, because element at index $2^{k-1}$ is lesser then $X$ and element at index $2^k$ is greater than $X$.

Let's have an example,

![enter image description here](https://github.com/KingsGambitLab/Lecture_Notes/blob/master/articles/Akash%20Articles/md/Images/Exponential%20Search/1.jpg)

![enter image description here](https://github.com/KingsGambitLab/Lecture_Notes/blob/master/articles/Akash%20Articles/md/Images/Exponential%20Search/2.jpg)

![enter image description here](https://github.com/KingsGambitLab/Lecture_Notes/blob/master/articles/Akash%20Articles/md/Images/Exponential%20Search/3.jpg)

![enter image description here](https://github.com/KingsGambitLab/Lecture_Notes/blob/master/articles/Akash%20Articles/md/Images/Exponential%20Search/4.jpg)
![enter image description here](https://github.com/KingsGambitLab/Lecture_Notes/blob/master/articles/Akash%20Articles/md/Images/Exponential%20Search/5.jpg)

Now, Let's see the implementation.

```cpp
#include <bits/stdc++.h>
using namespace std;

int binary_search(vector<int>& list, int l, int r, int key)
{
    while(l <= r) {
        int mid = (l + r)/2;
        if(list[mid] == key)
            return mid;
        else if(list[mid] < key)
            l = mid + 1;
        else
            r = mid - 1;
    }
    
    return -1; 
}

int exponential_search(vector<int>& list, int key)
{
    int size = list.size() - 1;
    
    if(list[0] == key)
        return 0;

    // Finding range
    int bound = 1;
    while (bound < size && list[bound] < key)
        bound *= 2;
    
    return binary_search(list, bound/2, min(bound + 1, size), key);
}

int main()
{
    vector<int> list{3,5,6,7,8,10,17,20,24,27};
    int key = 10;
    
    cout << exponential_search(list,key) << endl;
    
    return 0;
}
```
**Note:** It is assumed that the elements in the list are unique.

### Quiz Time
Can you figure out the time complexity in terms of the index($i$) of the element?
**Answer:** $\mathcal{O}(log(i))$
**Explanation:** In order to find an index of form $2^k$, such that $list[2^k] > key$, we are running the while loop $\lceil\log{i}\rceil$ times, that is $\mathcal{O}(\log(i))$ time complexity.

After that, binary search on the range $2^{\log(i)-1}$ to $2^{\log(i)}$, that is interval of size $2^{\log(i)-1}$, takes $\log{(2^{\log(i)-1})}$ comparisons, which leads to $\mathcal{O}(\log(i))$ complexity.

So the overall time complexity is $\mathcal{O}(\log(i))$.

Can you figure out the best time complexity?
**Answer:** $\mathcal{O}(1)$
**Explanation:** When the element we are searching is at the first index.

### When to use exponential search?
Worst case time-complexity of exponential search is $\mathcal{O}(\log(n))$. 

When the element is near to the front of the list, exponential search performs better than binary search for the case of a very large list. 

For an example purpose, Let say for below array we are searching for 3,`[1,3,4,6,10,14,18,20,22,23,25,28,30,33,36,39,40,41]`. Then it is easy to see that binary search certainly takes more comparisons then exponential search.

**Therefore, when the list size is very big and it looks like the element is too far from the end of the list, use exponential search.**