Create Exponential_Search.md

articles/Akash Articles/md/Exponential_Search.md

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+## 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 that this?
+
+Apparantely, 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 index-range for a given element?
+
+As the name of the algorithm suggests, we are going to use exponentiation. We find an index of 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 <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 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.**