## 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://lh3.googleusercontent.com/5NjJTuBXtIAxPlwv9ZoFG9SLNRN56E_iC4zXIZ1pDu4ipFRkuAlG001c9STTgTfSOFtVTlIu11mM) ![enter image description here](https://lh3.googleusercontent.com/hIiFWzFhtnV4RKNsIMPdcwxryes9DYMkX9HM6Kp4lE0nx8aJKy1U4DO6FEQauxukZnoKG54zAMpN) ![enter image description here](https://lh3.googleusercontent.com/bGRaeI3f4bucSeNqT1JFEFsc8HK8nh8o8sikRhV6QcaOar4qAYV9ITN3pD5TV6CEQsavzO0K4wEh) ![enter image description here](https://lh3.googleusercontent.com/UeoCFXPtEcQ6ucUCkn1FbOST2PI7DjiLXnDHlK-EoX5hXDj4Zj12mjVvvlYZt2UpEFYzCiZrg_vo) ![enter image description here](https://lh3.googleusercontent.com/xVS0E4OEd_aLFb2HpcFbqqpgoQQIlq2kak07LDuCFiYO7nyzmJvBlmWs6EL4dISlR36BhqcVBwD_) Now, Let's see the implementation. ```cpp #include using namespace std; int binary_search(vector& 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& 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 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.**