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 Nth pass.
- 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).
- 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) = (3N + 3 * (-1N)) / 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> |
14763
Time Complexity: O(log n)
Auxiliary Space: O(1)