Find the GCD of N Fibonacci Numbers with given Indices
Last Updated :
02 May, 2025
Given indices of N Fibonacci numbers. The task is to find the GCD of the Fibonacci numbers present at the given indices.
The first few Fibonacci numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89...
Note: The indices start from zero. That is, 0th Fibonacci number = 0.
Examples:
Input: Indices = {2, 3, 4, 5}
Output: GCD of the fibonacci numbers = 1
Input: Indices = {3, 6, 9}
Output: GCD of the fibonacci numbers = 2
Brute Force Approach:
- Initialize an array to store the Fibonacci numbers.
- Calculate the maximum index in the given list of indices.
- Calculate all the Fibonacci numbers up to the maximum index using a loop.
- Extract the Fibonacci numbers at the given indices from the array.
- Compute the GCD of the extracted Fibonacci numbers using a loop and the Euclidean algorithm.
- Return the GCD as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to compute the nth Fibonacci number
int getFib(int n)
{
if (n <= 1) return n;
return getFib(n-1) + getFib(n-2);
}
// Function to find the GCD of Fibonacci numbers
// at given indices using brute force approach
int findGCD(int indices[], int n)
{
int maxIndex = *max_element(indices, indices+n);
vector<int> fib(maxIndex+1);
// Compute all Fibonacci numbers up to maxIndex
for (int i = 0; i <= maxIndex; i++)
fib[i] = getFib(i);
// Compute the GCD of Fibonacci numbers at given indices
int gcd = fib[indices[0]];
for (int i = 1; i < n; i++)
gcd = __gcd(gcd, fib[indices[i]]);
return gcd;
}
// Driver code
int main()
{
int indices[] = { 3, 6, 9 };
int n = sizeof(indices)/sizeof(indices[0]);
cout << findGCD(indices, n) << endl;
return 0;
}
import java.util.Arrays;
import java.util.Vector;
public class GFG {
// Function to compute the nth Fibonacci number
public static int getFib(int n) {
if (n <= 1) return n;
return getFib(n-1) + getFib(n-2);
}
// Function to find the GCD of Fibonacci numbers
// at given indices using brute force approach
public static int findGCD(int[] indices, int n) {
int maxIndex = Arrays.stream(indices).max().getAsInt();
Vector<Integer> fib = new Vector<Integer>(maxIndex+1);
// Compute all Fibonacci numbers up to maxIndex
for (int i = 0; i <= maxIndex; i++)
fib.add(getFib(i));
// Compute the GCD of Fibonacci numbers at given indices
int gcd = fib.get(indices[0]);
for (int i = 1; i < n; i++)
gcd = gcd(gcd, fib.get(indices[i]));
return gcd;
}
// Function to compute the GCD of two numbers
public static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver code
public static void main(String[] args) {
int[] indices = { 3, 6, 9 };
int n = indices.length;
System.out.println(findGCD(indices, n));
}
}
import math
# Function to compute the nth Fibonacci number
def get_fib(n):
if n <= 1:
return n
return get_fib(n-1) + get_fib(n-2)
# Function to find the GCD of Fibonacci numbers
# at given indices using brute force approach
def find_gcd(indices):
max_index = max(indices)
fib = [0] * (max_index+1)
# Compute all Fibonacci numbers up to max_index
for i in range(max_index+1):
fib[i] = get_fib(i)
# Compute the GCD of Fibonacci numbers at given indices
gcd = fib[indices[0]]
for i in range(1, len(indices)):
gcd = math.gcd(gcd, fib[indices[i]])
return gcd
# Driver code
indices = [3, 6, 9]
print(find_gcd(indices))
using System;
using System.Collections.Generic;
public class GFG
{
// Function to compute the nth Fibonacci number
public static int GetFib(int n)
{
if (n <= 1) return n;
return GetFib(n - 1) + GetFib(n - 2);
}
// Function to find the GCD of Fibonacci numbers
// at given indices using the brute force approach
public static int FindGCD(int[] indices)
{
int maxIndex = indices[0];
for (int i = 1; i < indices.Length; i++)
{
if (indices[i] > maxIndex)
maxIndex = indices[i];
}
// Compute all Fibonacci numbers up to maxIndex
List<int> fib = new List<int>();
for (int i = 0; i <= maxIndex; i++)
fib.Add(GetFib(i));
// Compute the GCD of Fibonacci numbers at given indices
int gcd = fib[indices[0]];
for (int i = 1; i < indices.Length; i++)
gcd = GCD(gcd, fib[indices[i]]);
return gcd;
}
// Function to compute the Greatest Common Divisor (GCD) of two numbers
public static int GCD(int a, int b)
{
while (b != 0)
{
int temp = b;
b = a % b;
a = temp;
}
return a;
}
// Driver code
public static void Main(string[] args)
{
int[] indices = { 3, 6, 9 };
Console.WriteLine(FindGCD(indices));
}
}
// Function to compute the nth Fibonacci number
function getFib(n)
{
if (n <= 1) return n;
return getFib(n-1) + getFib(n-2);
}
// Function to computr gcd
function GCD(a, b)
{
if(b == 0)
return a;
return GCD(b, a % b);
}
// Function to find the GCD of Fibonacci numbers
// at given indices using brute force approach
function findGCD(indices, n)
{
let maxIndex = Math.max(...indices);
let fib = new Array(maxIndex+1);
// Compute all Fibonacci numbers up to maxIndex
for (let i = 0; i <= maxIndex; i++)
fib[i] = getFib(i);
// Compute the GCD of Fibonacci numbers at given indices
let gcd = fib[indices[0]];
for (let i = 1; i < n; i++)
gcd = GCD(gcd, fib[indices[i]]);
return gcd;
}
// Driver code
let indices = [ 3, 6, 9 ];
let n = indices.length;
document.write(findGCD(indices, n));
Time Complexity: O(k * n), where k is the number of indices and n is the maximum index in the given array.
Space Complexity: O(maxIndex), where maxIndex is the maximum index in the given array of indices.
Efficient Approach: An efficient approach is to use the property:
GCD(Fib(M), Fib(N)) = Fib(GCD(M, N))
The idea is to calculate the GCD of all the indices and then find the Fibonacci number at the index gcd_1( where gcd_1 is the GCD of the given indices).
Below is the implementation of the above approach:
C++
// C++ program to Find the GCD of N Fibonacci
// Numbers with given Indices
#include <bits/stdc++.h>
using namespace std;
// Function to return n'th
// Fibonacci number
int getFib(int n)
{
/* Declare an array to store Fibonacci numbers. */
int f[n + 2]; // 1 extra to handle case, n = 0
int i;
// 0th and 1st number of the series
// are 0 and 1
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++) {
// Add the previous 2 numbers in the series
// and store it
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
// Function to Find the GCD of N Fibonacci
// Numbers with given Indices
int find(int arr[], int n)
{
int gcd_1 = 0;
// find the gcd of the indices
for (int i = 0; i < n; i++) {
gcd_1 = __gcd(gcd_1, arr[i]);
}
// find the fibonacci number at
// index gcd_1
return getFib(gcd_1);
}
// Driver code
int main()
{
int indices[] = { 3, 6, 9 };
int N = sizeof(indices) / sizeof(int);
cout << find(indices, N);
return 0;
}
// Java program to Find the GCD of N Fibonacci
// Numbers with given Indices
import java.io.*;
// Function to return n'th
// Fibonacci number
public class GFG {
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a-b, b);
return __gcd(a, b-a);
}
static int getFib(int n)
{
/* Declare an array to store Fibonacci numbers. */
int f[] = new int[n + 2];
// 1 extra to handle case, n = 0
int i;
// 0th and 1st number of the series
// are 0 and 1
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++) {
// Add the previous 2 numbers in the series
// and store it
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
// Function to Find the GCD of N Fibonacci
// Numbers with given Indices
static int find(int arr[], int n)
{
int gcd_1 = 0;
// find the gcd of the indices
for (int i = 0; i < n; i++) {
gcd_1 = __gcd(gcd_1, arr[i]);
}
// find the fibonacci number at
// index gcd_1
return getFib(gcd_1);
}
// Driver code
public static void main (String[] args) {
int indices[] = { 3, 6, 9 };
int N = indices.length;
System.out.println( find(indices, N));
}
}
# Python program to Find the
# GCD of N Fibonacci Numbers
# with given Indices
from math import *
# Function to return n'th
# Fibonacci number
def getFib(n) :
# Declare an array to store
# Fibonacci numbers.
f = [0] * (n + 2) # 1 extra to handle case, n = 0
# 0th and 1st number of the
# series are 0 and 1
f[0], f[1] = 0, 1
# Add the previous 2 numbers
# in the series and store it
for i in range(2, n + 1) :
f[i] = f[i - 1] + f[i - 2]
return f[n]
# Function to Find the GCD of N Fibonacci
# Numbers with given Indices
def find(arr, n) :
gcd_1 = 0
# find the gcd of the indices
for i in range(n) :
gcd_1 = gcd(gcd_1, arr[i])
# find the fibonacci number
# at index gcd_1
return getFib(gcd_1)
# Driver code
if __name__ == "__main__" :
indices = [3, 6, 9]
N = len(indices)
print(find(indices, N))
# This code is contributed by ANKITRAI1
// C# program to Find the GCD
// of N Fibonacci Numbers with
// given Indices
using System;
// Function to return n'th
// Fibonacci number
class GFG
{
// Recursive function to
// return gcd of a and b
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
static int getFib(int n)
{
/* Declare an array to
store Fibonacci numbers. */
int []f = new int[n + 2];
// 1 extra to handle case, n = 0
int i;
// 0th and 1st number of
// the series are 0 and 1
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
// Add the previous 2 numbers
// in the series and store it
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
// Function to Find the GCD
// of N Fibonacci Numbers
// with given Indices
static int find(int []arr, int n)
{
int gcd_1 = 0;
// find the gcd of the indices
for (int i = 0; i < n; i++)
{
gcd_1 = __gcd(gcd_1, arr[i]);
}
// find the fibonacci number
// at index gcd_1
return getFib(gcd_1);
}
// Driver code
public static void Main ()
{
int []indices = { 3, 6, 9 };
int N = indices.Length;
Console.WriteLine(find(indices, N));
}
}
// This code is contributed
// by Shashank
<script>
// javascript program to
// Find the GCD of N Fibonacci
// Numbers with given Indices
// Function to return n'th
// Fibonacci number
// Recursive function to return gcd of a and b
function __gcd(a , b) {
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
function getFib(n) {
/* Declare an array to store Fibonacci numbers. */
var f = Array(n + 2).fill(0);
// 1 extra to handle case, n = 0
var i;
// 0th and 1st number of the series
// are 0 and 1
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++) {
// Add the previous 2 numbers in the series
// and store it
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
// Function to Find the GCD of N Fibonacci
// Numbers with given Indices
function find(arr , n) {
var gcd_1 = 0;
// find the gcd of the indices
for (i = 0; i < n; i++) {
gcd_1 = __gcd(gcd_1, arr[i]);
}
// find the fibonacci number at
// index gcd_1
return getFib(gcd_1);
}
// Driver code
var indices = [ 3, 6, 9 ];
var N = indices.length;
document.write(find(indices, N));
// This code contributed by gauravrajput1
</script>
<?php
// PHP program to Find the GCD of
// N Fibonacci Numbers with given
// Indices
// Function to return n'th
// Fibonacci number
function gcd($a, $b)
{
return $b ? gcd($b, $a % $b) : $a;
}
function getFib($n)
{
/* Declare an array to store
Fibonacci numbers. */
// 1 extra to handle case, n = 0
$f = array_fill(0, ($n + 2), NULL);
// 0th and 1st number of the
// series are 0 and 1
$f[0] = 0;
$f[1] = 1;
for ($i = 2; $i <= $n; $i++)
{
// Add the previous 2 numbers
// in the series and store it
$f[$i] = $f[$i - 1] + $f[$i - 2];
}
return $f[$n];
}
// Function to Find the GCD of N Fibonacci
// Numbers with given Indices
function find(&$arr, $n)
{
$gcd_1 = 0;
// find the gcd of the indices
for ($i = 0; $i < $n; $i++)
{
$gcd_1 = gcd($gcd_1, $arr[$i]);
}
// find the fibonacci number
// at index gcd_1
return getFib($gcd_1);
}
// Driver code
$indices = array(3, 6, 9 );
$N = sizeof($indices);
echo find($indices, $N);
// This code is contributed
// by ChitraNayal
?>
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
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