Exponential Search also known as finger search, searches for an element in a sorted array by jumping `2^i` elements every iteration where i represents the value of loop control variable, and then verifying if the search element is present between last jump and the current jump.

## Complexity Worst Case

O(log(N)) Often confused because of the name, the algorithm is named so not because of the time complexity. The name arises as a result of the algorithm jumping elements with steps equal to exponents of 2

## How it works

1. Jump the array `2^i` elements at a time searching for the condition `Array[2^(i-1)] < valueWanted < Array[2^i]`. If `2^i` is greater than the lenght of array, then set the upper bound to the length of the array.
2. Do a binary search between `Array[2^(i-1)]` and `Array[2^i]`

## The Code

``````// 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;
}

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;
}``````
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}