Find resulting Colour by Combination of given 3 colours in Array

Given a String of three Colours(G, B, Y) as input, the task is to print the resultant combined color formed according to the rule given below: 

// Rules for colour combination

• Blue(B) * Green(G) = Yellow(Y)
• Yellow(Y) * Blue(B) = Green(G)
• Green(G) * Yellow(Y) = Blue(B)

Examples:

Input: str = “GBYGB”
Output B

Input: str = “BYB”
Output Y

Approach: This problem can be solved as follows: 

1. Get the input string.
2. Compare each alphabet with its adjacent characters.
3. Use the above condition to determine the combination.
4. print the output combination.

Below is the implementation of the above approach:  

C++

// C++ program to find the
// resultant colour combination

#include <iostream>
using namespace std;

// Function to return Colour Combination
char Colour_Combination(string s)
{
    char temp = s[0];

    for (int i = 1; i < s.length(); i++) {
        if (temp != s[i]) {

            // Check for B * G = Y
            if ((temp == 'B' || temp == 'G')
                && (s[i] == 'G' || s[i] == 'B'))
                temp = 'Y';

            // Check for B * Y = G
            else if ((temp == 'B' || temp == 'Y')
                     && (s[i] == 'Y' || s[i] == 'B'))
                temp = 'G';

            // Check for Y * G = B
            else
                temp = 'B';
        }
    }
    return temp;
}

// Driver Code
int main(int argc, char** argv)
{
    string s = "GBYGB";

    cout << Colour_Combination(s);
}

Java

// Java program to find the
// resultant colour combination
class GfG
{

    // Function to return Colour Combination
    static char Colour_Combination(String s)
    {
        char temp = s.charAt(0);

        for (int i = 1; i < s.length(); i++)
        {
            if (temp != s.charAt(i))
            {

                // Check for B * G = Y
                if ((temp == 'B' || temp == 'G')
                            && (s.charAt(i) == 'G'
                            || s.charAt(i) == 'B'))
                    temp = 'Y';

                // Check for B * Y = G
                else if ((temp == 'B' || temp == 'Y')
                                && (s.charAt(i) == 'Y'
                                || s.charAt(i) == 'B'))
                    temp = 'G';

                // Check for Y * G = B
                else
                    temp = 'B';
            }
        }
        return temp;
    }

    // Driver code
    public static void main(String []args)
    {
        String s = "GBYGB";
        System.out.println(Colour_Combination(s));
    }
}

// This code is contributed by Rituraj Jain

Python3

# Python 3 program to find the
# resultant colour combination

# Function to return Colour Combination
def Colour_Combination(s):
    temp = s[0]

    for i in range(1, len(s), 1):
        if (temp != s[i]):

            # Check for B * G = Y
            if ((temp == 'B' or temp == 'G') and
                (s[i] == 'G' or s[i] == 'B')):
                temp = 'Y'

            # Check for B * Y = G
            elif ((temp == 'B' or temp == 'Y') and
                  (s[i] == 'Y' or s[i] == 'B')):
                temp = 'G'

            # Check for Y * G = B
            else:
                temp = 'B'
    return temp

# Driver Code
if __name__ == '__main__':
    s = "GBYGB"

    print(Colour_Combination(s))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find the
// resultant colour combination

using System;

class GFG
{

    // Function to return Colour Combination
    static char Colour_Combination(string s)
    {
        char temp = s[0];

        for (int i = 1; i < s.Length; i++)
        {
            if (temp != s[i])
            {

                // Check for B * G = Y
                if ((temp == 'B' || temp == 'G')
                    && (s[i] == 'G' || s[i] == 'B'))
                    temp = 'Y';

                // Check for B * Y = G
                else if ((temp == 'B' || temp == 'Y')
                        && (s[i] == 'Y' || s[i] == 'B'))
                    temp = 'G';

                // Check for Y * G = B
                else
                    temp = 'B';
            }
        }
        return temp;
    }

    // Driver Code
    static void Main()
    {
        string s = "GBYGB";

        Console.WriteLine(Colour_Combination(s));
    }
}

// This code is contributed by Ryuga

PHP

<?php
// PHP program to find the
// resultant colour combination

// Function to return Colour Combination
function Colour_Combination($s)
{
    $temp = $s[0];

    for ($i = 1; $i < strlen($s); $i++)
    {
        if ($temp != $s[$i])
        {

            // Check for B * G = Y
            if (($temp == 'B' || $temp == 'G') &&
                ($s[$i] == 'G' || $s[$i] == 'B'))
                $temp = 'Y';

            // Check for B * Y = G
            else if (($temp == 'B' || $temp == 'Y') &&
                     ($s[$i] == 'Y' || $s[$i] == 'B'))
                $temp = 'G';

            // Check for Y * G = B
            else
                $temp = 'B';
        }
    }
    return $temp;
}

// Driver Code
$s = "GBYGB";

echo Colour_Combination($s);

// This code is contributed by ita_c
?>

Javascript

<script>

// Javascript program to find the
// resultant colour combination

// Function to return Colour Combination
function Colour_Combination(s)
{
    let temp = s[0];

    for(let i = 1; i < s.length; i++)
    {
        if (temp != s[i])
        {

            // Check for B * G = Y
            if ((temp == 'B' || temp == 'G') &&
                (s[i] == 'G' || s[i] == 'B'))
                temp = 'Y';

            // Check for B * Y = G
            else if ((temp == 'B' || temp == 'Y') &&
                     (s[i] == 'Y' || s[i] == 'B'))
                temp = 'G';

            // Check for Y * G = B
            else
                temp = 'B';
        }
    }
    return temp;
}

// Driver Code
let s = "GBYGB";

document.write(Colour_Combination(s));

// This code is contributed by Surbhi Tyagi.

</script>

Output

B

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1)