Number of ways to color the balls (Ball Coloring)

Given N balls. All of them are initially uncolored. You have to color the balls with two colors RED and BLUE such that there can be at most 2 positions where a RED ball is touching BLUE ball or vice versa. You have to color all the balls. Find the number of ways in which balls can be colored.

Examples:

Input: n = 1
Output: 2
Explanation: Possible ways to color are {{R}, {B}}. So, the answer is 2.

Input: n = 2
Output: 4
Explanation: Possible ways to color are {{RB}, {BR}, {RR}, {BB}}. So, the answer is 4.

Approach: To solve the problem, follow the below idea:

We can color all the balls in 3 ways.

1. Zero red and blue balls are touching: We can color N balls in this way in 2 ways. For example: RRR , BBB is valid coloring for 3 balls.
2. Red and Blue balls are touching at one place: We can color N balls in this way in 2*(N-1) ways. For example: RRB, RBB, BBR, BRR for 3 balls
3. Red and Blue balls are touching at one place: In this case, we have to change the color of balls in 2 positions. For N balls we can choose 1st position in (N-1) ways and 2nd position in (N-2) ways. So we can color all the balls in (N-2)*(N-1) ways . Because at least First and Last Ball should have same color to satisfy the condition. For example: RBR, BRB

So total number of ways is ANS = 2 + 2 * (N-1) + (N-2) * (N-1) = 2 + N * (N-1)

Step-by-step algorithm:

• Input the value of N.
• Find the product as prod = N * (N – 1).
• Return prod + 2.

Below is the implementation of the approach:

C++

#include <iostream>
using namespace std;

unsigned long long int noOfWays(unsigned long long int n)
{
    // code here
    unsigned long long int prod = n * (n - 1);
    return prod + 2;
}

int main()
{
    // Input
    unsigned long long int n = 4;
    cout << noOfWays(n);
    return 0;
}

Java

public class Main {
    static long noOfWays(long n) {
        // code here
        long prod = n * (n - 1);
        return prod + 2;
    }

    public static void main(String[] args) {
        // Input
        long n = 4;
        System.out.println(noOfWays(n));
    }
}

Python3

# Python Implementation

def no_of_ways(n):
    prod = n * (n - 1)
    return prod + 2

n = 4
print(no_of_ways(n))

# This code is contributed by Tapesh(tapeshdu420)

C#

using System;

public class Program
{
    static long NoOfWays(long n)
    {
        // code here
        long prod = n * (n - 1);
        return prod + 2;
    }

    public static void Main(string[] args)
    {
        // Input
        long n = 4;
        Console.WriteLine(NoOfWays(n));
    }
}

Javascript

function noOfWays(n) {
    // code here
    let prod = n * (n - 1);
    return prod + 2;
}

// Input
let n = 4;
console.log(noOfWays(n));

Output

14

Time Complexity: O(1)
Auxiliary Space: O(1)