# Count of ordered triplets (R, G, B) in a given original string

Given a string S of size N, the task is to count all possible triplets in the given string which consists of only 3 colours (R)Red, (G)Green and (B)Blue in the order (R, G, B).

Examples:

Input: S = “RRGB”
Output: 2
Explanation: There are two triplets of RGB in the given string:

• R at index 0, G at index 2 and B at index 3 forms one triplet of RGB.
• R at index 1, G at index 2 and B at index 3 forms the second triplet of RGB.

Input: S = “GBR”
Output: 0
Explanation: No triplets exists.

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

Approach:

1. Count the number of B(Blue) in the given string and store the value in a Blue_Count variable.
2. Initialize Red_Count = 0.
3. Iterate over all character of string from left to right.
4. If the current character is (R)red, increase the Red_Count.
5. If current character is (B)blue, decrease Blue_Count.
6. If the current character is G(green) add Red_Count * Blue_Count in the result.

Below is the implementation of the above approach.

## C++

 `// C++ code for the above program ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// function to count the ` `// ordered triplets (R, G, B) ` `int` `countTriplets(string color) ` `{ ` `    ``int` `result = 0, Blue_Count = 0; ` `    ``int` `Red_Count = 0; ` ` `  `    ``// count the B(blue) colour ` `    ``for` `(``char` `c : color) { ` `        ``if` `(c == ``'B'``) ` `            ``Blue_Count++; ` `    ``} ` ` `  `    ``for` `(``char` `c : color) { ` `        ``if` `(c == ``'B'``) ` `            ``Blue_Count--; ` `        ``if` `(c == ``'R'``) ` `            ``Red_Count++; ` `        ``if` `(c == ``'G'``) ` `            ``result += Red_Count * Blue_Count; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``string color = ``"RRGGBBRGGBB"``; ` `    ``cout << countTriplets(color); ` `    ``return` `0; ` `} `

## Java

 `// Java code for the above program ` `class` `GFG{ ` `         `  `// function to count the ` `// ordered triplets (R, G, B) ` `static` `int` `countTriplets(String color) ` `{ ` `    ``int` `result = ``0``, Blue_Count = ``0``; ` `    ``int` `Red_Count = ``0``; ` `    ``int` `len = color.length(); ` `    ``int` `i; ` `     `  `    ``// count the B(blue) colour ` `    ``for` `(i = ``0``; i < len ; i++) ` `    ``{ ` `        ``if` `(color.charAt(i) == ``'B'``) ` `            ``Blue_Count++; ` `    ``} ` ` `  `    ``for` `(i = ``0``; i < len ; i++) ` `    ``{ ` `        ``if` `(color.charAt(i) == ``'B'``) ` `            ``Blue_Count--; ` `        ``if` `(color.charAt(i) == ``'R'``) ` `            ``Red_Count++; ` `        ``if` `(color.charAt(i) == ``'G'``) ` `            ``result += Red_Count * Blue_Count; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``String color = ``"RRGGBBRGGBB"``; ` `    ``System.out.println(countTriplets(color)); ` `} ` ` `  `} ` ` `  `// This code is contributed by AnkitRai01  `

## Python3

 `# Python3 code for the above program  ` ` `  `# function to count the  ` `# ordered triplets (R, G, B)  ` `def` `countTriplets(color) : ` ` `  `    ``result ``=` `0``; Blue_Count ``=` `0``;  ` `    ``Red_Count ``=` `0``;  ` ` `  `    ``# count the B(blue) colour  ` `    ``for` `c ``in` `color : ` `        ``if` `(c ``=``=` `'B'``) : ` `            ``Blue_Count ``+``=` `1``;  ` ` `  `    ``for` `c ``in` `color :  ` `        ``if` `(c ``=``=` `'B'``) : ` `            ``Blue_Count ``-``=` `1``;  ` `             `  `        ``if` `(c ``=``=` `'R'``) : ` `            ``Red_Count ``+``=` `1``;  ` `             `  `        ``if` `(c ``=``=` `'G'``) : ` `            ``result ``+``=` `Red_Count ``*` `Blue_Count;  ` `     `  `    ``return` `result;  ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``color ``=` `"RRGGBBRGGBB"``;  ` `    ``print``(countTriplets(color));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# code for the above program ` `using` `System; ` `class` `GFG{ ` `         `  `// function to count the ` `// ordered triplets (R, G, B) ` `static` `int` `countTriplets(String color) ` `{ ` `    ``int` `result = 0, Blue_Count = 0; ` `    ``int` `Red_Count = 0; ` `    ``int` `len = color.Length; ` `    ``int` `i; ` ` `  `    ``// count the B(blue) colour ` `    ``for` `(i = 0; i < len ; i++) ` `    ``{ ` `        ``if` `(color[i] == ``'B'``) ` `            ``Blue_Count++; ` `    ``} ` ` `  `    ``for` `(i = 0; i < len ; i++) ` `    ``{ ` `        ``if` `(color[i] == ``'B'``) ` `            ``Blue_Count--; ` `        ``if` `(color[i] == ``'R'``) ` `            ``Red_Count++; ` `        ``if` `(color[i] == ``'G'``) ` `            ``result += Red_Count * Blue_Count; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``string` `color = ``"RRGGBBRGGBB"``; ` `    ``Console.WriteLine(countTriplets(color)); ` `} ` `} ` ` `  `// This code is contributed by AnkitRai01  `

Output:

```28
```

Time Complexity: O(N)
Auxillary Space Cpmplexity: O(1)

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