/**
 * Exponential Search
 *
 * The algorithm consists of two stages. The first stage determines a
 * range in which the search key would reside if it were in the list.
 * In the second stage, a binary search is performed on this range.
 *
 *
 *
 */

function binarySearch (arr, x, floor, ceiling) {
  // Middle index
  const mid = Math.floor((floor + ceiling) / 2)

  // If value is at the mid position return this position
  if (arr[mid] === x) {
    return mid
  }

  if (floor > ceiling) return -1

  // If the middle element is great than the value
  // search the left part of the array
  if (arr[mid] > value) {
    return binarySearch(arr, value, floor, mid - 1)
    // If the middle element is lower than the value
    // search the right part of the array
  } else {
    return binarySearch(arr, value, mid + 1, ceiling)
  }
}

function exponentialSearch (arr, length, value) {
  // If value is the first element of the array return this position
  if (arr[0] === value) {
    return 0
  }

  // Find range for binary search
  let i = 1
  while (i < length && arr[i] <= value) {
    i = i * 2
  }

  // Call binary search for the range found above
  return binarySearch(arr, value, i / 2, Math.min(i, length))
}

const arr = [2, 3, 4, 10, 40, 65, 78, 100]
const value = 78
const result = exponentialSearch(arr, arr.length, value)

if (result < 0) {
  console.log('Element not found')
} else {
  console.log('Element found at position :' + result)
}

Prerequisites

• Binary Search algorithm

Problem Statement

Given a sorted array of n elements, write a function to search for the index of a given element (target)

Approach

• Search for the range within which the target is included increasing index by powers of 2
• If this range exists in array apply the Binary Search algorithm over it
• Else return -1

Example

arr = [1, 2, 3, 4, 5, 6, 7, ... 998, 999, 1_000]

target = 998
index = 0
1. SEARCHING FOR THE RANGE
index = 1, 2, 4, 8, 16, 32, 64, ..., 512, ..., 1_024
after 10 iteration we have the index at 1_024 and outside of the array 
2. BINARY SEARCH
Now we can apply the binary search on the subarray from 512 and 1_000.

Note: we apply the Binary Search from 512 to 1_000 because at i = 2^10 = 1_024 the array is finisced and the target number is less than the latest index of the array ( 1_000 ).

Time Complexity

worst case: O(log *i*) where *i* = index (position) of the target

best case: O(*1*)

Complexity Explanation

• The complexity of the first part of the algorithm is O( log i ) because if i is the position of the target in the array, after doubling the search index ⌈log(i)⌉ times, the algorithm will be at a search index that is greater than or equal to i. We can write 2^⌈log(i)⌉ >= i
• The complexity of the second part of the algorithm also is O ( log i ) because that is a simple Binary Search. The Binary Search complexity ( as explained here ) is O( n ) where n is the length of the array. In the Exponential Search, the length of the array on which the algorithm is applied is 2^i - 2^(i-1), put into words it means '( the length of the array from start to i ) - ( the part of array skipped until the previous iteration )'. Is simple verify that 2^i - 2^(i-1) = 2^(i-1)

After this detailed explanation we can say that the the complexity of the Exponential Search is:

O(log i) + O(log i) = 2O(log i) = O(log i)

Binary Search vs Exponential Search

Let's take a look at this comparison with a less theoretical example. Imagine we have an array with1_000_000 elements and we want to search an element that is in the 4th position. It's easy to see that:

• The Binary Search start from the middle of the array and arrive to the 4th position after many iterations
• The Exponential Search arrive at the 4th index after only 2 iterations

Code Implementation Links

• C++
• JavaScript