Fibonacci Search

Given a sorted array arr[] of size n and an element x to be searched in it. Return index of x if it is present in array else return -1. 
Examples: 

Input:  arr[] = {2, 3, 4, 10, 40}, x = 10
Output:  3
Element x is present at index 3.

Input:  arr[] = {2, 3, 4, 10, 40}, x = 11
Output:  -1
Element x is not present.

Fibonacci Search is a comparison-based technique that uses Fibonacci numbers to search an element in a sorted array.
Similarities with Binary Search:

1. Works for sorted arrays
2. A Divide and Conquer Algorithm.
3. Has Log n time complexity.

Differences with Binary Search: 

1. Fibonacci Search divides given array into unequal parts
2. Binary Search uses a division operator to divide range. Fibonacci Search doesn’t use /, but uses + and -. The division operator may be costly on some CPUs.
3. Fibonacci Search examines relatively closer elements in subsequent steps. So when the input array is big that cannot fit in CPU cache or even in RAM, Fibonacci Search can be useful.

Background:
Fibonacci Numbers are recursively defined as F(n) = F(n-1) + F(n-2), F(0) = 0, F(1) = 1. First few Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …
Observations:
Below observation is used for range elimination, and hence for the O(log(n)) complexity.  

F(n - 2) ≈ (1/3)*F(n) and 
F(n - 1) ≈ (2/3)*F(n).

Algorithm:
Let the searched element be x.
The idea is to first find the smallest Fibonacci number that is greater than or equal to the length of the given array. Let the found Fibonacci number be fib (m’th Fibonacci number). We use (m-2)’th Fibonacci number as the index (If it is a valid index). Let (m-2)’th Fibonacci Number be i, we compare arr[i] with x, if x is same, we return i. Else if x is greater, we recur for subarray after i, else we recur for subarray before i.
Below is the complete algorithm 
Let arr[0..n-1] be the input array and the element to be searched be x.  

1. Find the smallest Fibonacci Number greater than or equal to n. Let this number be fibM [m’th Fibonacci Number]. Let the two Fibonacci numbers preceding it be fibMm1 [(m-1)’th Fibonacci Number] and fibMm2 [(m-2)’th Fibonacci Number].
2. While the array has elements to be inspected: 
   1. Compare x with the last element of the range covered by fibMm2
   2. If x matches, return index
   3. Else If x is less than the element, move the three Fibonacci variables two Fibonacci down, indicating elimination of approximately rear two-third of the remaining array.
   4. Else x is greater than the element, move the three Fibonacci variables one Fibonacci down. Reset offset to index. Together these indicate the elimination of approximately front one-third of the remaining array.
3. Since there might be a single element remaining for comparison, check if fibMm1 is 1. If Yes, compare x with that remaining element. If match, return index.

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to find minimum of two elements
int min(int x, int y) { return (x <= y) ? x : y; }

/* Returns index of x if present,  else returns -1 */
int fibMonaccianSearch(int arr[], int x, int n)
{
    /* Initialize fibonacci numbers */
    int fibMMm2 = 0; // (m-2)'th Fibonacci No.
    int fibMMm1 = 1; // (m-1)'th Fibonacci No.
    int fibM = fibMMm2 + fibMMm1; // m'th Fibonacci

    /* fibM is going to store the smallest Fibonacci
       Number greater than or equal to n */
    while (fibM < n) {
        fibMMm2 = fibMMm1;
        fibMMm1 = fibM;
        fibM = fibMMm2 + fibMMm1;
    }

    // Marks the eliminated range from front
    int offset = -1;

    /* while there are elements to be inspected. Note that
       we compare arr[fibMm2] with x. When fibM becomes 1,
       fibMm2 becomes 0 */
    while (fibM > 1) {
        // Check if fibMm2 is a valid location
        int i = min(offset + fibMMm2, n - 1);

        /* If x is greater than the value at index fibMm2,
           cut the subarray array from offset to i */
        if (arr[i] < x) {
            fibM = fibMMm1;
            fibMMm1 = fibMMm2;
            fibMMm2 = fibM - fibMMm1;
            offset = i;
        }

        /* If x is greater than the value at index fibMm2,
           cut the subarray after i+1  */
        else if (arr[i] > x) {
            fibM = fibMMm2;
            fibMMm1 = fibMMm1 - fibMMm2;
            fibMMm2 = fibM - fibMMm1;
        }

        /* element found. return index */
        else
            return i;
    }

    /* comparing the last element with x */
    if (fibMMm1 && arr[offset + 1] == x)
        return offset + 1;

    /*element not found. return -1 */
    return -1;
}

// Driver Code
int main()
{
    int arr[]
        = { 10, 22, 35, 40, 45, 50, 80, 82, 85, 90, 100,235};
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 235;
      int ind = fibMonaccianSearch(arr, x, n);
  if(ind>=0)
    cout << "Found at index: " << ind;
  else
    cout << x << " isn't present in the array";

    return 0;
}

// This code is contributed by code_hunt.

C

// C program for Fibonacci Search
#include <stdio.h>

// Utility function to find minimum of two elements
int min(int x, int y) { return (x <= y) ? x : y; }

/* Returns index of x if present,  else returns -1 */
int fibMonaccianSearch(int arr[], int x, int n)
{
    /* Initialize fibonacci numbers */
    int fibMMm2 = 0; // (m-2)'th Fibonacci No.
    int fibMMm1 = 1; // (m-1)'th Fibonacci No.
    int fibM = fibMMm2 + fibMMm1; // m'th Fibonacci

    /* fibM is going to store the smallest Fibonacci
       Number greater than or equal to n */
    while (fibM < n) {
        fibMMm2 = fibMMm1;
        fibMMm1 = fibM;
        fibM = fibMMm2 + fibMMm1;
    }

    // Marks the eliminated range from front
    int offset = -1;

    /* while there are elements to be inspected. Note that
       we compare arr[fibMm2] with x. When fibM becomes 1,
       fibMm2 becomes 0 */
    while (fibM > 1) {
        // Check if fibMm2 is a valid location
        int i = min(offset + fibMMm2, n - 1);

        /* If x is greater than the value at index fibMm2,
           cut the subarray array from offset to i */
        if (arr[i] < x) {
            fibM = fibMMm1;
            fibMMm1 = fibMMm2;
            fibMMm2 = fibM - fibMMm1;
            offset = i;
        }

        /* If x is greater than the value at index fibMm2,
           cut the subarray after i+1  */
        else if (arr[i] > x) {
            fibM = fibMMm2;
            fibMMm1 = fibMMm1 - fibMMm2;
            fibMMm2 = fibM - fibMMm1;
        }

        /* element found. return index */
        else
            return i;
    }

    /* comparing the last element with x */
    if (fibMMm1 && arr[offset + 1] == x)
        return offset + 1;

    /*element not found. return -1 */
    return -1;
}

/* driver function */
int main(void)
{
    int arr[]
        = { 10, 22, 35, 40, 45, 50, 80, 82, 85, 90, 100,235};
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 235;
      int ind = fibMonaccianSearch(arr, x, n);
  if(ind>=0)
    printf("Found at index: %d",ind);
  else
    printf("%d isn't present in the array",x);
    return 0;
}

Java

// Java program for Fibonacci Search
import java.util.*;

class Fibonacci {
    // Utility function to find minimum
    // of two elements
    public static int min(int x, int y)
    {
        return (x <= y) ? x : y;
    }

    /* Returns index of x if present, else returns -1 */
    public static int fibMonaccianSearch(int arr[], int x,
                                         int n)
    {
        /* Initialize fibonacci numbers */
        int fibMMm2 = 0; // (m-2)'th Fibonacci No.
        int fibMMm1 = 1; // (m-1)'th Fibonacci No.
        int fibM = fibMMm2 + fibMMm1; // m'th Fibonacci

        /* fibM is going to store the smallest
        Fibonacci Number greater than or equal to n */
        while (fibM < n) {
            fibMMm2 = fibMMm1;
            fibMMm1 = fibM;
            fibM = fibMMm2 + fibMMm1;
        }

        // Marks the eliminated range from front
        int offset = -1;

        /* while there are elements to be inspected.
        Note that we compare arr[fibMm2] with x.
        When fibM becomes 1, fibMm2 becomes 0 */
        while (fibM > 1) {
            // Check if fibMm2 is a valid location
            int i = min(offset + fibMMm2, n - 1);

            /* If x is greater than the value at
            index fibMm2, cut the subarray array
            from offset to i */
            if (arr[i] < x) {
                fibM = fibMMm1;
                fibMMm1 = fibMMm2;
                fibMMm2 = fibM - fibMMm1;
                offset = i;
            }

            /* If x is less than the value at index
            fibMm2, cut the subarray after i+1 */
            else if (arr[i] > x) {
                fibM = fibMMm2;
                fibMMm1 = fibMMm1 - fibMMm2;
                fibMMm2 = fibM - fibMMm1;
            }

            /* element found. return index */
            else
                return i;
        }

        /* comparing the last element with x */
        if (fibMMm1 == 1 && arr[n-1] == x)
            return n-1;

        /*element not found. return -1 */
        return -1;
    }

    // driver code
    public static void main(String[] args)
    {
        int arr[] = { 10, 22, 35, 40, 45, 50,
                      80, 82, 85, 90, 100,235};
        int n = 12;
        int x = 235;
      int ind = fibMonaccianSearch(arr, x, n);
        if(ind>=0)
        System.out.print("Found at index: "
                         +ind);
      else
        System.out.print(x+" isn't present in the array");
    }
}

// This code is contributed by rishabh_jain

Python3

# Python3 program for Fibonacci search.
from bisect import bisect_left

# Returns index of x if present,  else
# returns -1


def fibMonaccianSearch(arr, x, n):

    # Initialize fibonacci numbers
    fibMMm2 = 0  # (m-2)'th Fibonacci No.
    fibMMm1 = 1  # (m-1)'th Fibonacci No.
    fibM = fibMMm2 + fibMMm1  # m'th Fibonacci

    # fibM is going to store the smallest
    # Fibonacci Number greater than or equal to n
    while (fibM < n):
        fibMMm2 = fibMMm1
        fibMMm1 = fibM
        fibM = fibMMm2 + fibMMm1

    # Marks the eliminated range from front
    offset = -1

    # while there are elements to be inspected.
    # Note that we compare arr[fibMm2] with x.
    # When fibM becomes 1, fibMm2 becomes 0
    while (fibM > 1):

        # Check if fibMm2 is a valid location
        i = min(offset+fibMMm2, n-1)

        # If x is greater than the value at
        # index fibMm2, cut the subarray array
        # from offset to i
        if (arr[i] < x):
            fibM = fibMMm1
            fibMMm1 = fibMMm2
            fibMMm2 = fibM - fibMMm1
            offset = i

        # If x is less than the value at
        # index fibMm2, cut the subarray
        # after i+1
        elif (arr[i] > x):
            fibM = fibMMm2
            fibMMm1 = fibMMm1 - fibMMm2
            fibMMm2 = fibM - fibMMm1

        # element found. return index
        else:
            return i

    # comparing the last element with x */
    if(fibMMm1 and arr[n-1] == x):
        return n-1

    # element not found. return -1
    return -1


# Driver Code
arr = [10, 22, 35, 40, 45, 50,
       80, 82, 85, 90, 100,235]
n = len(arr)
x = 235
ind = fibMonaccianSearch(arr, x, n)
if ind>=0:
  print("Found at index:",ind)
else:
  print(x,"isn't present in the array");

# This code is contributed by rishabh_jain

C#

// C# program for Fibonacci Search
using System;

class GFG {

    // Utility function to find minimum
    // of two elements
    public static int min(int x, int y)
    {
        return (x <= y) ? x : y;
    }

    /* Returns index of x if present, else returns -1 */
    public static int fibMonaccianSearch(int[] arr, int x,
                                         int n)
    {
        /* Initialize fibonacci numbers */
        int fibMMm2 = 0; // (m-2)'th Fibonacci No.
        int fibMMm1 = 1; // (m-1)'th Fibonacci No.
        int fibM = fibMMm2 + fibMMm1; // m'th Fibonacci

        /* fibM is going to store the smallest
        Fibonacci Number greater than or equal to n */
        while (fibM < n) {
            fibMMm2 = fibMMm1;
            fibMMm1 = fibM;
            fibM = fibMMm2 + fibMMm1;
        }

        // Marks the eliminated range from front
        int offset = -1;

        /* while there are elements to be inspected.
        Note that we compare arr[fibMm2] with x.
        When fibM becomes 1, fibMm2 becomes 0 */
        while (fibM > 1) {
            // Check if fibMm2 is a valid location
            int i = min(offset + fibMMm2, n - 1);

            /* If x is greater than the value at
            index fibMm2, cut the subarray array
            from offset to i */
            if (arr[i] < x) {
                fibM = fibMMm1;
                fibMMm1 = fibMMm2;
                fibMMm2 = fibM - fibMMm1;
                offset = i;
            }

            /* If x is less than the value at index
            fibMm2, cut the subarray after i+1 */
            else if (arr[i] > x) {
                fibM = fibMMm2;
                fibMMm1 = fibMMm1 - fibMMm2;
                fibMMm2 = fibM - fibMMm1;
            }

            /* element found. return index */
            else
                return i;
        }

        /* comparing the last element with x */
        if (fibMMm1 == 1 && arr[n-1] == x)
            return n-1;

        /*element not found. return -1 */
        return -1;
    }

    // driver code
    public static void Main()
    {
        int[] arr = { 10, 22, 35, 40, 45, 50,
                      80, 82, 85, 90, 100,235 };
        int n = 12;
        int x = 235;
          int ind = fibMonaccianSearch(arr, x, n);
          if(ind>=0)
        Console.Write("Found at index: "+ind);
          else
        Console.Write(x+" isn't present in the array");
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP program for Fibonacci Search

/* Returns index of x if present, else returns -1 */
function fibMonaccianSearch($arr, $x, $n)
{
    /* Initialize fibonacci numbers */
    $fibMMm2 = 0; // (m-2)'th Fibonacci No.
    $fibMMm1 = 1; // (m-1)'th Fibonacci No.
    $fibM = $fibMMm2 + $fibMMm1; // m'th Fibonacci

    /* fibM is going to store the smallest Fibonacci
    Number greater than or equal to n */
    while ($fibM < $n)
    {
        $fibMMm2 = $fibMMm1;
        $fibMMm1 = $fibM;
        $fibM = $fibMMm2 + $fibMMm1;
    }

    // Marks the eliminated range from front
    $offset = -1;

    /* while there are elements to be inspected. Note that
    we compare arr[fibMm2] with x. When fibM becomes 1,
    fibMm2 becomes 0 */
    while ($fibM > 1)
    {
        // Check if fibMm2 is a valid location
        $i = min($offset+$fibMMm2, $n-1);

        /* If x is greater than the value at index fibMm2,
        cut the subarray array from offset to i */
        if ($arr[$i] < $x)
        {
            $fibM = $fibMMm1;
            $fibMMm1 = $fibMMm2;
            $fibMMm2 = $fibM - $fibMMm1;
            $offset = $i;
        }

        /* If x is less than the value at index fibMm2,
        cut the subarray after i+1 */
        else if ($arr[$i] > $x)
        {
            $fibM = $fibMMm2;
            $fibMMm1 = $fibMMm1 - $fibMMm2;
            $fibMMm2 = $fibM - $fibMMm1;
        }

        /* element found. return index */
        else return $i;
    }

    /* comparing the last element with x */
    if($fibMMm1 && $arr[$n-1] == $x)return $n-1;

    /*element not found. return -1 */
    return -1;
}

/* driver code */
    $arr = array(10, 22, 35, 40, 45, 50, 80, 82,85, 90, 100,235);
    $n = count($arr);
    $x = 235;
    $ind = fibMonaccianSearch($arr, $x, $n);
    if($ind>=0)
    printf("Found at index: ".$ind);
    else
    printf($x." isn't present in the array");

// This code is contributed by mits
?>

Javascript

<script>

// Javascript program for Fibonacci Search

/* Returns index of x if present, else returns -1 */
function fibMonaccianSearch(arr, x, n)
{
    /* Initialize fibonacci numbers */
    let fibMMm2 = 0; // (m-2)'th Fibonacci No.
    let fibMMm1 = 1; // (m-1)'th Fibonacci No.
    let fibM = fibMMm2 + fibMMm1; // m'th Fibonacci

    /* fibM is going to store the smallest Fibonacci
    Number greater than or equal to n */
    while (fibM < n)
    {
        fibMMm2 = fibMMm1;
        fibMMm1 = fibM;
        fibM = fibMMm2 + fibMMm1;
    }

    // Marks the eliminated range from front
    let offset = -1;

    /* while there are elements to be inspected. Note that
    we compare arr[fibMm2] with x. When fibM becomes 1,
    fibMm2 becomes 0 */

    while (fibM > 1)
    {
        // Check if fibMm2 is a valid location
        let i = Math.min(offset + fibMMm2, n-1);

        /* If x is greater than the value at index fibMm2,
        cut the subarray array from offset to i */
        if (arr[i] < x)
        {
            fibM = fibMMm1;
            fibMMm1 = fibMMm2;
            fibMMm2 = fibM - fibMMm1;
            offset = i;
        }

        /* If x is less than the value at index fibMm2,
        cut the subarray after i+1 */
        else if (arr[i] > x)
        {
            fibM = fibMMm2;
            fibMMm1 = fibMMm1 - fibMMm2;
            fibMMm2 = fibM - fibMMm1;
        }

        /* element found. return index */
        else return i;
    }

    /* comparing the last element with x */
    if(fibMMm1 && arr[n-1] == x){
      return n-1
    }

    /*element not found. return -1 */
    return -1;
}

/* driver code */
    let arr = [10, 22, 35, 40, 45, 50, 80, 82,85, 90, 100,235];
    let n = arr.length;
    let x = 235;
    let ind = fibMonaccianSearch(arr, x, n);
    if(ind>=0){
       document.write("Found at index: " + ind);
    }else{
       document.write(x + " isn't present in the array");
    }

// This code is contributed by _saurabh_jaiswal

</script>

Output

Found at index: 11

Illustration:
Let us understand the algorithm with the below example:  

Illustration assumption: 1-based indexing. Target element x is 85. Length of array n = 11.
Smallest Fibonacci number greater than or equal to 11 is 13. As per our illustration, fibMm2 = 5, fibMm1 = 8, and fibM = 13.
Another implementation detail is the offset variable (zero-initialized). It marks the range that has been eliminated, starting from the front. We will update it from time to time.
Now since the offset value is an index and all indices including it and below it have been eliminated, it only makes sense to add something to it. Since fibMm2 marks approximately one-third of our array, as well as the indices it marks, are sure to be valid ones, we can add fibMm2 to offset and check the element at index i = min(offset + fibMm2, n). 

Visualization:

Time Complexity analysis:
The worst-case will occur when we have our target in the larger (2/3) fraction of the array, as we proceed to find it. In other words, we are eliminating the smaller (1/3) fraction of the array every time. We call once for n, then for(2/3) n, then for (4/9) n, and henceforth.
Consider that: 

Auxiliary Space: O(1)

References:
https://en.wikipedia.org/wiki/Fibonacci_search_technique

This article is contributed by Yash Varyani. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Approach 2: Iterative implementation

Fibonacci Search is a searching algorithm used to find the position of an element in a sorted array. The basic idea behind Fibonacci Search is to use Fibonacci numbers to determine the split points in the array and perform binary search on the appropriate subarray.

Here’s a Python implementation of Fibonacci Search using an iterative approach:

C++

// C++ code for above approach
#include <iostream>
using namespace std;

int fibonacciSearch(int arr[], int n, int x) {
if (n == 0) {
return -1;
}
// Initialize Fibonacci numbers
int fib1 = 0, fib2 = 1, fib3 = fib1 + fib2;

// Find the smallest Fibonacci number greater than or equal to n
while (fib3 < n) {
    fib1 = fib2;
    fib2 = fib3;
    fib3 = fib1 + fib2;
}

// Initialize variables for the current and previous split points
int offset = -1;
while (fib3 > 1) {
    int i = min(offset + fib2, n-1);

    // If x is greater than the value at index i, move the split point to the right
    if (arr[i] < x) {
        fib3 = fib2;
        fib2 = fib1;
        fib1 = fib3 - fib2;
        offset = i;
    }

    // If x is less than the value at index i, move the split point to the left
    else if (arr[i] > x) {
        fib3 = fib1;
        fib2 = fib2 - fib1;
        fib1 = fib3 - fib2;
    }

    // If x is equal to the value at index i, return the index
    else {
        return i;
    }
}

// If x is not found in the array, return -1
if (fib2 == 1 && arr[offset+1] == x) {
    return offset + 1;
} else {
    return -1;
}
}

// Driver code
int main() {
int arr[] = {10, 22, 35, 40, 45, 50, 80, 82, 85, 90, 100, 235};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 235;
int ind = fibonacciSearch(arr, n, x);
if (ind >= 0) {
cout << "Found at index: " << ind << endl;
} else {
cout << x << " isn't present in the array" << endl;
}
return 0;
}

Java

import java.util.*;

class Main {
  public static int fibonacciSearch(int[] arr, int x) {
    int n = arr.length;
    if (n == 0) {
      return -1;
    }

    // Initialize Fibonacci numbers
    int fib1 = 0, fib2 = 1, fib3 = fib1 + fib2;

    // Find the smallest Fibonacci number greater than or equal to n
    while (fib3 < n) {
      fib1 = fib2;
      fib2 = fib3;
      fib3 = fib1 + fib2;
    }

    // Initialize variables for the current and previous split points
    int offset = -1;
    while (fib3 > 1) {
      int i = Math.min(offset + fib2, n-1);

      // If x is greater than the value at index i, move the split point to the right
      if (arr[i] < x) {
        fib3 = fib2;
        fib2 = fib1;
        fib1 = fib3 - fib2;
        offset = i;
      }

      // If x is less than the value at index i, move the split point to the left
      else if (arr[i] > x) {
        fib3 = fib1;
        fib2 = fib2 - fib1;
        fib1 = fib3 - fib2;
      }

      // If x is equal to the value at index i, return the index
      else {
        return i;
      }
    }

    // If x is not found in the array, return -1
    if (fib2 == 1 && arr[offset+1] == x) {
      return offset + 1;
    } else {
      return -1;
    }
  }

  public static void main(String[] args) {
    int[] arr = {10, 22, 35, 40, 45, 50, 80, 82, 85, 90, 100, 235};
    int n = arr.length;
    int x = 235;
    int ind = fibonacciSearch(arr, x);
    if (ind >= 0) {
      System.out.println("Found at index: " + ind);
    } else {
      System.out.println(x + " isn't present in the array");
    }
  }
}

Python3

def fibonacci_search(arr, x):
    n = len(arr)
    if n == 0:
        return -1

    # Initialize Fibonacci numbers
    fib1, fib2 = 0, 1
    fib3 = fib1 + fib2

    # Find the smallest Fibonacci number greater than or equal to n
    while fib3 < n:
        fib1, fib2 = fib2, fib3
        fib3 = fib1 + fib2

    # Initialize variables for the current and previous split points
    offset = -1
    while fib3 > 1:
        i = min(offset + fib2, n-1)

        # If x is greater than the value at index i, move the split point to the right
        if arr[i] < x:
            fib3 = fib2
            fib2 = fib1
            fib1 = fib3 - fib2
            offset = i

        # If x is less than the value at index i, move the split point to the left
        elif arr[i] > x:
            fib3 = fib1
            fib2 = fib2 - fib1
            fib1 = fib3 - fib2

        # If x is equal to the value at index i, return the index
        else:
            return i

    # If x is not found in the array, return -1
    if fib2 == 1 and arr[offset+1] == x:
        return offset + 1
    else:
        return -1




# Driver Code
arr = [10, 22, 35, 40, 45, 50,
    80, 82, 85, 90, 100,235]
n = len(arr)
x = 235
ind = fibonacci_search(arr, x)
if ind>=0:
    print("Found at index:",ind)
else:
    print(x,"isn't present in the array");

# This code is contributed by ajay singh

C#

using System;

class GFG {
    public static int FibonacciSearch(int[] arr, int x) {
        int n = arr.Length;
        if (n == 0) {
            return -1;
        }

        // Initialize Fibonacci numbers
        int fib1 = 0, fib2 = 1, fib3 = fib1 + fib2;

        // Find the smallest Fibonacci number greater than or equal to n
        while (fib3 < n) {
            fib1 = fib2;
            fib2 = fib3;
            fib3 = fib1 + fib2;
        }

        // Initialize variables for the current and previous split points
        int offset = -1;
        while (fib3 > 1) {
            int i = Math.Min(offset + fib2, n - 1);

            // If x is greater than the value at index i, move the split point to the right
            if (arr[i] < x) {
                fib3 = fib2;
                fib2 = fib1;
                fib1 = fib3 - fib2;
                offset = i;
            }

            // If x is less than the value at index i, move the split point to the left
            else if (arr[i] > x) {
                fib3 = fib1;
                fib2 = fib2 - fib1;
                fib1 = fib3 - fib2;
            }

            // If x is equal to the value at index i, return the index
            else {
                return i;
            }
        }

        // If x is not found in the array, return -1
        if (fib2 == 1 && arr[offset + 1] == x) {
            return offset + 1;
        } else {
            return -1;
        }
    }

      // Driver code
    public static void Main(string[] args) {
        int[] arr = { 10, 22, 35, 40, 45, 50, 80, 82, 85, 90, 100, 235 };
        int n = arr.Length;
        int x = 235;

          // function call
        int ind = FibonacciSearch(arr, x);
        if (ind >= 0) {
            Console.WriteLine("Found at index: " + ind);
        } else {
            Console.WriteLine(x + " isn't present in the array");
        }
    }
}

Javascript

// Javascript equivalent
function fibonacci_search(arr, x) {
    var n = arr.length;
    if (n === 0) {
        return -1;
    }

    // Initialize Fibonacci numbers
    var fib1 = 0;
    var fib2 = 1;
    var fib3 = fib1 + fib2;

    // Find the smallest Fibonacci number greater than or equal to n
    while (fib3 < n) {
        fib1 = fib2;
        fib2 = fib3;
        fib3 = fib1 + fib2;
    }

    // Initialize variables for the current and previous split points
    var offset = -1;
    while (fib3 > 1) {
        var i = Math.min(offset + fib2, n - 1);

        // If x is greater than the value at index i,
        // move the split point to the right
        if (arr[i] < x) {
            fib3 = fib2;
            fib2 = fib1;
            fib1 = fib3 - fib2;
            offset = i;

            // If x is less than the value at index i,
            // move the split point to the left
        } else if (arr[i] > x) {
            fib3 = fib1;
            fib2 = fib2 - fib1;
            fib1 = fib3 - fib2;

            // If x is equal to the value at index i, return the index
        } else {
            return i;
        }
    }

    // If x is not found in the array, return -1
    if (fib2 === 1 && arr[offset + 1] === x) {
        return offset + 1;
    } else {
        return -1;
    }
}

// Driver Code
var arr = [10, 22, 35, 40, 45, 50,
    80, 82, 85, 90, 100,235];
var n = arr.length;
var x = 235;
var ind = fibonacci_search(arr, x);
if (ind >= 0) {
    console.log("Found at index:", ind);
} else {
    console.log(x,"isn't present in the array");
}

Output

Found at index: 11

The time complexity of Fibonacci Search is O(log n), where n is the length of the input array.

This is because at each iteration of the algorithm, the search range is reduced by a factor of approximately 1/φ, where φ is the golden ratio (φ ≈ 1.618). The number of iterations required to reduce the search range to a single element is approximately logₑ(n), where logₑ denotes the natural logarithm.

Since each iteration of Fibonacci Search requires constant time, the overall time complexity of the algorithm is O(log n). This makes Fibonacci Search a faster algorithm than linear search, but slower than binary search and other logarithmic search algorithms such as interpolation search and exponential search.

Auxiliary Space: O(1)