Exponential Search

• Difficulty Level : Easy
• Last Updated : 06 Oct, 2021

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed, linear search, binary search for this problem.
Exponential search involves two steps:  

1. Find range where element is present
2. 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)
Given below are the implementations of above steps.

• C++
• C
• Java
• Python
• C#
• PHP
• Javascript

// 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 occurrence of

// x in array

int exponentialSearch(int arr[], int n, int x)

{

    // If x is present at first 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-1), 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)? cout <<"Element is not present in array"

                 : cout <<"Element is present at index " << result;

   return 0;

}

// this code is contributed by shivanisinghss2110

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: 

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