The name of this searching algorithm may be misleading as it works in O(Log n) time. The name comes from the way it searches an element.

Given a sorted array an element x to be searched, find position of x in the array. Input: arr[] = {10, 20, 40, 45, 55} x = 45 Output: Element found at index 3 Input: arr[] = {10, 15, 25, 45, 55} x = 15 Output: Element found at index 1

We have discussed, linear search, binary search for this problem.

Exponential search involves two steps:

- Find range where element is present
- Do Binary Search in above found range.

**How to find the range where element may be present?**

The idea is to start with subarray size 1 compare its last element with x, then try size 2, then 4 and so on until last element of a subarray is not greater.

Once we find an index i (after repeated doubling of i), we know that the element must be present between i/2 and i (Why i/2? because we could not find a greater value in previous iteration)

Below is implementation of above steps.

## C++

// C++ program to find an element x in a // sorted array using Exponential search. #include <bits/stdc++.h> using namespace std; int binarySearch(int arr[], int, int, int); // Returns position of first ocurrence of // x in array int exponentialSearch(int arr[], int n, int x) { // If x is present at firt location itself if (arr[0] == x) return 0; // Find range for binary search by // repeated doubling int i = 1; while (i < n && arr[i] <= x) i = i*2; // Call binary search for the found range. return binarySearch(arr, i/2, min(i, n), x); } // A recursive binary search function. It returns // location of x in given array arr[l..r] is // present, otherwise -1 int binarySearch(int arr[], int l, int r, int x) { if (r >= l) { int mid = l + (r - l)/2; // If the element is present at the middle // itself if (arr[mid] == x) return mid; // If element is smaller than mid, then it // can only be present n left subarray if (arr[mid] > x) return binarySearch(arr, l, mid-1, x); // Else the element can only be present // in right subarray return binarySearch(arr, mid+1, r, x); } // We reach here when element is not present // in array return -1; } // Driver code int main(void) { int arr[] = {2, 3, 4, 10, 40}; int n = sizeof(arr)/ sizeof(arr[0]); int x = 10; int result = exponentialSearch(arr, n, x); (result == -1)? printf("Element is not present in array") : printf("Element is present at index %d", result); return 0; }

## Java

// Java program to find an element x in a // sorted array using Exponential search. import java.util.Arrays; class Test { // Returns position of first ocurrence of // x in array static int exponentialSearch(int arr[], int n, int x) { // If x is present at firt location itself if (arr[0] == x) return 0; // Find range for binary search by // repeated doubling int i = 1; while (i < n && arr[i] <= x) i = i*2; // Call binary search for the found range. return Arrays.binarySearch(arr, i/2, Math.min(i, n), x); } // Driver method public static void main(String args[]) { int arr[] = {2, 3, 4, 10, 40}; int x = 10; int result = exponentialSearch(arr, arr.length, x); System.out.println((result < 0) ? "Element is not present in array" : "Element is present at index " + result); } }

Output :

Element is present at index 3

**Time Complexity : **O(Log n)

**Auxiliary Space :** The above implementation of Binary Search is recursive and requires O()Log n) space. With iterative Binary Search, we need only O(1) space.

**Applications of Exponential Search:**

- Exponential Binary Search is particularly useful for unbounded searches, where size of array is infinite. Please refer Unbounded Binary Search for an example.
- It works better than Binary Search for bounded arrays also when the element to be searched is closer to the first element.

**Reference:**

https://en.wikipedia.org/wiki/Exponential_search

This article is contributed by **Pankaj Sharma**. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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