Fibonacci problem (Value of Fib(N)*Fib(N) – Fib(N-1) * Fib(N+1))

Given a positive integer N, the task is to find Fib(N)² – (Fib(N-1) * Fib(N+1)) where Fib(N) returns the Nth Fibonacci number.
Examples:

Input: N = 3 
Output: 1 
Fib(3) * Fib(3) – Fib(2) * Fib(4) = 4 – 3 = 1
Input: N = 2 
Output: -1 
Fib(2) * Fib(2) – Fib(1) * Fib(3) = 1 – 2 = -1 

Approach: This question can be approached by first trying out a few test cases. Let’s take a few examples: 

For N = 1 : Fib(1) * Fib(1) – Fib(0) * Fib(2) = 1 – 0 = 1 
For N = 2 : Fib(2) * Fib(2) – Fib(1) * Fib(3) = 1 – 2 = -1 
For N = 3 : Fib(3) * Fib(3) – Fib(2) * Fib(4) = 4 – 3 = 1 
For N = 4 : Fib(4) * Fib(4) – Fib(3) * Fib(5) = 9 – 10 = -1 
We observe here that when N is even then the answer will be -1 and when the N is odd then the answer will be 1. 

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

int getResult(int n)
{
    if (n & 1)
        return 1;
    return -1;
}

// Driver code
int main()
{
    int n = 3;
    cout << getResult(n);
}

Java

//Java implementation of the approach

import java.io.*;

class GFG {

static int getResult(int n)
{
    if ((n & 1)>0)
        return 1;
    return -1;
}

// Driver code
    public static void main (String[] args) {

    int n = 3;
    System.out.println(getResult(n));
    }
//This code is contributed by @Tushil.
}

Python3

# Python 3 implementation of
# the approach
def getResult(n):

    if n & 1:
        return 1
    return -1

# Driver code
n = 3
print(getResult(n))

# This code is contributed
# by Shrikant13

C#

//C# implementation of the approach

using System;

class GFG {

static int getResult(int n)
{
    if ((n & 1)>0)
        return 1;
    return -1;
}

// Driver code
    public static void Main () {

    int n = 3;
    Console.WriteLine(getResult(n));
    }
//This code is contributed by anuj_67..
}

PHP

<?php
// PHP implementation of the approach

function getResult($n)
{
    if ($n & 1)
        return 1;
    return -1;
}

// Driver code
$n = 3;
echo getResult($n);

// This code is contributed by akt_mit
?>

Javascript

<script>
    // Javascript implementation of the approach

    function getResult(n)
    {
        if ((n & 1)>0)
            return 1;
        return -1;
    }

    let n = 3;
    document.write(getResult(n));

</script>

Output:

1

Time Complexity: O(1)

Auxiliary Space: O(1)