给定原始字符串中有序三元组（R、G、B）的计数

给定一个大小为N的字符串S ，任务是计算给定字符串所有可能的三元组，该字符串仅由 3 种颜色 (R) 红色、(G) 绿色和 (B) 蓝色组成，顺序为(R, G, B) .
例子：

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. 
 

方法：

1. 计算给定字符串中 B(Blue) 的数量并将值存储在Blue_Count变量中。
2. 初始化Red_Count = 0。
3. 从左到右遍历字符串的所有字符。
4. 如果当前字符是 (R)red，则增加Red_Count 。
5. 如果当前字符是 (B)blue，则减少Blue_Count 。
6. 如果当前字符是 G(绿色)，则在结果中添加Red_Count * Blue_Count 。

下面是上述方法的实现。

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

Javascript

 
// Javascript code for the above program
 
// function to count the
// ordered triplets (R, G, B)
function countTriplets(color)
{
    var result = 0, Blue_Count = 0;
    var Red_Count = 0;
 
    var data = color.split('');
 
    // count the B(blue) colour
    for (var i=0;i<data.length;i++)
    {
        if (data[i] == 'B')
            Blue_Count++;
    }
 
    for (var i=0;i<data.length;i++){
        if (data[i] == 'B')
            Blue_Count--;
        if (data[i] == 'R')
            Red_Count++;
        if (data[i] == 'G')
            result += Red_Count * Blue_Count;
    }
    return result;
}
 
// Driver program
var color = "RRGGBBRGGBB";
document.write( countTriplets(color));
 

输出：

28

时间复杂度： O(N)
辅助空间复杂度： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live