In how many ways the ball will come back to the first boy after N turns

Four boys are playing a game with a ball. In each turn, the player (who has the ball currently) passes it to a different player randomly. Bob always starts the game. The task is to find in how many ways the ball will come back to Bob after N passes.
Examples:

Input: N = 3 
Output: 6 
Here are all the possible ways: 
Bob -> Boy1 -> Boy2 -> Bob 
Bob -> Boy1 -> Boy3 -> Bob 
Bob -> Boy2 -> Boy1 -> Bob 
Bob -> Boy2 -> Boy3 -> Bob 
Bob -> Boy3 -> Boy1 -> Bob 
Bob -> Boy3 -> Boy2 -> Bob
Input: N = 10 
Output: 14763 

Approach: Let the number of sequences that get back to Bob after N passes are P(N). There are two cases, either pass N – 2 is to Bob or it is not. Note that Bob can’t have the ball at (N – 1)^th pass because then he won’t have the ball at the N^th pass.

1. Case 1: If pass N – 2 is to Bob then the pass N – 1 can be to any of the other 3 boys. Thus, the number of such sequences is 3 * P(N – 2).
2. Case 2: If pass N – 2 is not to Bob then pass N – 1 is to one of the 2 boys other than Bob and the one who got the ball in hand. So, substitute Bob for the receiver of pass N – 1, and get a unique N – 1 sequence. So, the number of such sequences are 2 * P(N – 1).

Hence the recurrence relation will be P(N) = 2 * P(N – 1) + 3 * P(N – 2) where P(0) = 1 and P(1) = 0.
After solving the recurrence relation, P(N) = (3^N + 3 * (-1^N)) / 4
Below is the implementation of the above approach: 

C++

// Function to return the number of
// sequences that get back to Bob
#include <bits/stdc++.h>
using namespace std;

int numSeq(int n)
{
    return (pow(3, n) + 3 * pow(-1, n)) / 4;
}

// Driver code
int main()
{
    int N = 10;
    printf("%d", numSeq(N));
    return 0;
}

// This code is contributed by Mohit kumar

Java

// Function to return the number of
// sequences that get back to Bob
import java.util.*;

class GFG
{

static int numSeq(int n)
{
    return (int) ((Math.pow(3, n) + 3 *
                    Math.pow(-1, n)) / 4);
}

// Driver code
public static void main(String[] args)
{
    int N = 10;
    System.out.printf("%d", numSeq(N));
}
}

// This code is contributed by Rajput-Ji

Python3

# Function to return the number of
# sequences that get back to Bob
def numSeq(n):
    return (pow(3, n) + 3 * pow(-1, n))//4

# Driver code
N = 10
print(numSeq(N))

C#

// C# implementation of the above approach
using System;

// Function to return the number of
// sequences that get back to Bob
class GFG
{

    static int numSeq(int n)
    {
        return (int) ((Math.Pow(3, n) + 3 *
                       Math.Pow(-1, n)) / 4);
    }

    // Driver code
    public static void Main()
    {
        int N = 10;
        Console.WriteLine(numSeq(N));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Function to return the number of
// sequences that get back to Bob
function numSeq(n)
{
    return  Math.floor((Math.pow(3, n) + 3 * Math.pow(-1, n)) / 4);
}

// Driver code

    let N = 10;
    document.write(numSeq(N));

// This code is contributed by Surbhi Tyagi.

</script>

Output:

14763

Time Complexity: O(log n)

Auxiliary Space: O(1)