Exponential Search
/**
* 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, value, floor, ceiling) {
// Middle index
const mid = Math.floor((floor + ceiling) / 2)
// If value is at the mid position return this position
if (arr[mid] === value) {
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))
}
export { binarySearch, exponentialSearch }
// const arr = [2, 3, 4, 10, 40, 65, 78, 100]
// const value = 78
// const result = exponentialSearch(arr, arr.length, value)
About this Algorithm
Prerequisites
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 write2^⌈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 that2^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
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