# Find the resulting Colour Combination

Given a String of three Colours(G, B, Y) as input, the task is to print the resultant combined colour 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output:

```B
```

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