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;