Find F(n) when F(i) and F(j) of a sequence are given

Given five integers i, F_i, j, F_j and N. Where F_i and F_j are the i^th and j^th term of a sequence which follows the Fibonacci recurrence i.e. F_N = F_{N – 1} + F_{N – 2}. The task is to find the N^th term of the original sequence.
Examples:

Input: i = 3, F₃ = 5, j = -1, F_-1 = 4, N = 5 
Output: 12 
Fibonacci sequence can be reconstructed using known values: 
…, F_-1 = 4, F₀ = -1, F₁ = 3, F₂ = 2, F₃ = 5, F₄ = 7, F₅ = 12, …
Input: i = 0, F₀ = 1, j = 1, F₁ = 4, N = -2 
Output: -2 

Approach: Note that, if the two consecutive terms of the Fibonacci sequence are known then the N^th term can easily be determined. Assuming i < j, as per Fibonacci condition: 

  F_i+1 = 1*F_i+1 + 0*F_i
  F_i+2 = 1*F_i+1 + 1*F_i
  F_i+3 = F_i+2 + F_i+1 = 2*F_i+1 + 1*F_i
  F_i+4 = F_i+3 + F_i+2 = 3*F_i+1 + 2*F_i
  F_i+5 = F_i+4 + F_i+3 = 5*F_i+1 + 3*F_i
  .. .. .. 
and so on 

Note that, the coefficients of F_i+1 and F_i in the above set of equations are nothing but the terms of Standard Fibonacci Sequence.
So, considering the Standard Fibonacci sequence i.e. f₀ = 0, f₁ = 1, f₂ = 1, f₃ = 2, f₄ = 3, … ; we can generalize, the above set of equations (for k > 0) as: 

  F_i+k = f_k*F_i+1 + f_k-1*F_i  …(1)

Now consider, 

  k = j-i   …(2)

Now, substituting eq.(2) in eq.(1), we get: 

  F_j = f_j-i*F_i+1 + f_j-i-1*F_i

Hence, we can calculate F_i+1 from known values of F_i and F_j. Now that we know two consecutive terms of sequence F, we can easily reconstruct F and determine the value of F_N.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include<bits/stdc++.h>

using namespace std;

// Function to calculate kth Fibonacci number
// in the standard Fibonacci sequence
int fibonacci(int k)
{
    int a = 0, b = 1, c = 0;
    if( k == 0)
        return a;
    if (k == 1)
        return b;
    for(int i = 2; i < k + 1; i++)
    {
        c = a + b;
        a = b;
        b = c;
    }
    return c;
}

// Function to determine the value of F(n)
int determineFn(int i, int Fi, int j, int Fj, int n)
{
    if (j < i)
    {
        swap(i, j);
        swap(Fi, Fj);
    }

    // Find the value of F(i + 1)
    // from F(i) and F(j)
    int Fi1 = (Fj - fibonacci(j - i - 1) * Fi) /
                    fibonacci(j - i);

    // When n is equal to i
    int Fn = 0;
    if (n == i)
        Fn = Fi;

    // When n is greater than i
    else if (n > i)
    {
        int b = Fi;
        Fn = Fi1;
        n = n - 1;
        while (n != i)
        {
            n = n - 1;
            int a = b;
            b = Fn;
            Fn = a + b;
        }
    }

    // When n is smaller than i
    else
    {
        int a = Fi;
        int b = Fi1;
        while (n != i)
        {
            n = n + 1;
            Fn = b - a;
            b = a;
            a = Fn;
        }
    }
    return Fn;
}

// Driver code
int main()
{
    int i = 3;
    int Fi = 5;
    int j = -1;
    int Fj = 4;
    int n = 5;
    cout << (determineFn(i, Fi, j, Fj, n));
}

// This code is contributed by Mohit Kumar

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

    // Function to calculate kth Fibonacci number
    // in the standard Fibonacci sequence
    static int fibonacci(int k)
    {
        int a = 0, b = 1, c = 0;
        if (k == 0)
            return a;
        if (k == 1)
            return b;
        for (int i = 2; i < k + 1; i++)
        {
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }

    // Function to determine the value of F(n)
    static int determineFn(int i, int Fi,
                           int j, int Fj, int n)
    {
        if (j < i)
        {

            // Swap i, j
            i = i + j;
            j = i - j;
            i = i - j;

            // swap Fi, Fj
            Fi = Fi + Fj;
            Fj = Fi - Fj;
            Fi = Fi - Fj;
        }

        // Find the value of F(i + 1)
        // from F(i) and F(j)
        int Fi1 = (Fj - fibonacci(j - i - 1) * Fi) /
                        fibonacci(j - i);

        // When n is equal to i
        int Fn = 0;
        if (n == i)
            Fn = Fi;

        // When n is greater than i
        else if (n > i)
        {
            int b = Fi;
            Fn = Fi1;
            n = n - 1;
            while (n != i)
            {
                n = n - 1;
                int a = b;
                b = Fn;
                Fn = a + b;
            }
        }

        // When n is smaller than i
        else
        {
            int a = Fi;
            int b = Fi1;
            while (n != i)
            {
                n = n + 1;
                Fn = b - a;
                b = a;
                a = Fn;
            }
        }
        return Fn;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int i = 3;
        int Fi = 5;
        int j = -1;
        int Fj = 4;
        int n = 5;
        System.out.println(determineFn(i, Fi, j, Fj, n));
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of the approach

# Function to calculate kth Fibonacci number
# in the standard Fibonacci sequence
def fibonacci(k):
    a = 0
    b = 1
    if k == 0:
        return a
    if k == 1:
        return b
    for i in range(2, k + 1):
        c = a + b
        a = b
        b = c
    return c

# Function to determine
# the value of F(n)
def determineFn(i, Fi, j, Fj, n):
    if j<i:
        i, j = j, i
        Fi, Fj = Fj, Fi

    # Find the value of F(i + 1)
    # from F(i) and F(j)
    Fi1 = (Fj - fibonacci(j-i-1)*Fi)\
                      //fibonacci(j-i)

    # When n is equal to i
    if n == i:
        Fn = Fi

    # When n is greater than i
    elif n>i:
        b = Fi
        Fn = Fi1
        n = n - 1
        while n != i:
            n = n - 1
            a = b
            b = Fn
            Fn = a + b

    # When n is smaller than i
    else:
        a = Fi
        b = Fi1
        while n != i:
            n = n + 1
            Fn = b - a
            b = a
            a = Fn
    return Fn

# Driver code
if __name__ == '__main__':
    i = 3
    Fi = 5
    j = -1
    Fj = 4
    n = 5
    print(determineFn(i, Fi, j, Fj, n))

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to calculate kth Fibonacci number
    // in the standard Fibonacci sequence
    static int fibonacci(int k)
    {
        int a = 0, b = 1, c = 0;
        if (k == 0)
            return a;
        if (k == 1)
            return b;
        for (int i = 2; i < k + 1; i++)
        {
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }

    // Function to determine the value of F(n)
    static int determineFn(int i, int Fi,
                           int j, int Fj, int n)
    {
        if (j < i)
        {

            // Swap i, j
            i = i + j;
            j = i - j;
            i = i - j;

            // swap Fi, Fj
            Fi = Fi + Fj;
            Fj = Fi - Fj;
            Fi = Fi - Fj;
        }

        // Find the value of F(i + 1)
        // from F(i) and F(j)
        int Fi1 = (Fj - fibonacci(j - i - 1) * Fi) /
                        fibonacci(j - i);

        // When n is equal to i
        int Fn = 0;
        if (n == i)
            Fn = Fi;

        // When n is greater than i
        else if (n > i)
        {
            int b = Fi;
            Fn = Fi1;
            n = n - 1;
            while (n != i)
            {
                n = n - 1;
                int a = b;
                b = Fn;
                Fn = a + b;
            }
        }

        // When n is smaller than i
        else
        {
            int a = Fi;
            int b = Fi1;
            while (n != i)
            {
                n = n + 1;
                Fn = b - a;
                b = a;
                a = Fn;
            }
        }
        return Fn;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int i = 3;
        int Fi = 5;
        int j = -1;
        int Fj = 4;
        int n = 5;
        Console.WriteLine(determineFn(i, Fi, j, Fj, n));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript implementation of the approach

// Function to calculate kth Fibonacci number
// in the standard Fibonacci sequence
function fibonacci(k)
{
    let a = 0, b = 1, c = 0;
    if( k == 0)
        return a;
    if (k == 1)
        return b;
    for(let i = 2; i < k + 1; i++)
    {
        c = a + b;
        a = b;
        b = c;
    }
    return c;
}

// Function to determine the value of F(n)
function determineFn(i, Fi, j, Fj, n)
{
    if (j < i)
    {
        // Swap i, j
        i = i + j;
        j = i - j;
        i = i - j;

        // swap Fi, Fj
        Fi = Fi + Fj;
        Fj = Fi - Fj;
        Fi = Fi - Fj;
    }

    // Find the value of F(i + 1)
    // from F(i) and F(j)
    let Fi1 = parseInt((Fj - fibonacci(j - i - 1) * Fi) /
                    fibonacci(j - i));

    // When n is equal to i
    let Fn = 0;
    if (n == i)
        Fn = Fi;

    // When n is greater than i
    else if (n > i)
    {
        let b = Fi;
        Fn = Fi1;
        n = n - 1;
        while (n != i)
        {
            n = n - 1;
            let a = b;
            b = Fn;
            Fn = a + b;
        }
    }

    // When n is smaller than i
    else
    {
        let a = Fi;
        let b = Fi1;
        while (n != i)
        {
            n = n + 1;
            Fn = b - a;
            b = a;
            a = Fn;
        }
    }
    return Fn;
}

// Driver code
    let i = 3;
    let Fi = 5;
    let j = -1;
    let Fj = 4;
    let n = 5;
    document.write(determineFn(i, Fi, j, Fj, n));

</script>

Output:

12

Time Complexity: O(n)

Auxiliary Space: O(1)