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.

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++

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// C++ code for the above program
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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 

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Python3

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# 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

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C#

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// 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 

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Output:

28

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

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Improved By : AnkitRai01