Algorithms and Data Structures

Exponential Search - a very fast search Algorithm

Exponential Search - a very fast search Algorithm

Alessio Farinelli's photo
Alessio Farinelli
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4 min read

Prerequisite

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 : Note: we apply the Binary Search from 512 to 1_000 because at i = 2^10 = 1_024 the array is finished 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 with 1_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

My implementation : C

Others : C++, JavaScript

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