package com.thealgorithms.searches;

import com.thealgorithms.devutils.searches.SearchAlgorithm;
import java.util.Arrays;

/**
 * ExponentialSearch is an algorithm that efficiently finds the position of a target
 * value within a sorted array. It works by expanding the range to find the bounds
 * where the target might exist and then using binary search within that range.
 *
 * <p>
 * Worst-case time complexity: O(log n)
 * Best-case time complexity: O(1) when the element is found at the first position.
 * Average time complexity: O(log n)
 * Worst-case space complexity: O(1)
 * </p>
 *
 * <p>
 * Note: This algorithm requires that the input array be sorted.
 * </p>
 */
class ExponentialSearch implements SearchAlgorithm {

    /**
     * Finds the index of the specified key in a sorted array using exponential search.
     *
     * @param array The sorted array to search.
     * @param key The element to search for.
     * @param <T> The type of the elements in the array, which must be comparable.
     * @return The index of the key if found, otherwise -1.
     */
    @Override
    public <T extends Comparable<T>> int find(T[] array, T key) {
        if (array.length == 0) {
            return -1;
        }
        if (array[0].equals(key)) {
            return 0;
        }
        if (array[array.length - 1].equals(key)) {
            return array.length - 1;
        }

        int range = 1;
        while (range < array.length && array[range].compareTo(key) < 0) {
            range = range * 2;
        }

        // The candidate block is the inclusive index range [range / 2, range], so the
        // exclusive upper bound handed to binarySearch has to be range + 1.
        final int index = Arrays.binarySearch(array, range / 2, Math.min(range + 1, array.length), key);
        return index >= 0 ? index : -1;
    }
}

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