以“ geeks”开头并以“ for”结尾的子字符串数

给定一个由小写英文字母组成的字符串str ，任务是查找以“ geeks”开头和以“ for”结尾的子字符串数。

例子：

Input: str = “geeksforgeeksisforgeeks”
Output: 3
“geeksfor”, “geeksforgeeksisfor” and “geeksisfor”
are the only valid substrings.

Input: str = “geeksforgeeks”
Output: 1

天真的方法：首先将计数器设置为0，然后遍历字符串，并且每当遇到子字符串“ geeks”时，从下一个索引开始再次遍历字符串，并尝试找到子字符串“ for” 。如果显示“ for” ，则增加计数器并最终打印出来。

高效的方法：为子字符串“ geeks”和“ for”设置两个计数器，例如c1和c2。进行迭代时，每当遇到子字符串“ geeks”时，就递增c1；每当遇到“ for”时，将c2 = c2 + c1设置为1 。这是因为每次出现的“怪胎”都将使用当前找到的“ for”组成一个有效的子字符串。最后打印c2 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of required substrings
int countSubStr(string s, int n)
{
    int c1 = 0, c2 = 0;
  
    // For every index of the string
    for (int i = 0; i < n; i++) {
  
        // If the substring starting at
        // the current index is "geeks"
        if (s.substr(i, 5) == "geeks")
            c1++;
  
        // If the substring is "for"
        if (s.substr(i, 3) == "for")
            c2 = c2 + c1;
    }
  
    return c2;
}
  
// Driver code
int main()
{
    string s = "geeksforgeeksisforgeeks";
    int n = s.size();
  
    cout << countSubStr(s, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
    // Function to return the count
    // of required substrings
    static int countSubStr(String s, int n)
    {
        int c1 = 0, c2 = 0;
  
        // For every index of the string
        for (int i = 0; i < n; i++)
        {
  
            // If the substring starting at
            // the current index is "geeks"
            if (i < n - 5 && 
                "geeks".equals(s.substring(i, i + 5))) 
            {
                c1++;
            }
  
            // If the substring is "for"
            if (i < n - 3 && 
                "for".equals(s.substring(i, i + 3))) 
            {
                c2 = c2 + c1;
            }
        }
        return c2;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String s = "geeksforgeeksisforgeeks";
        int n = s.length();
        System.out.println(countSubStr(s, n));
    }
} 
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
  
# Function to return the count 
# of required substrings 
def countSubStr(s, n) : 
  
    c1 = 0; c2 = 0; 
  
    # For every index of the string 
    for i in range(n) :
  
        # If the substring starting at 
        # the current index is "geeks" 
        if (s[i : i + 5] == "geeks") :
            c1 += 1; 
  
        # If the substring is "for" 
        if (s[i :i+ 3] == "for") :
            c2 = c2 + c1; 
  
    return c2; 
  
# Driver code 
if __name__ == "__main__" : 
  
    s = "geeksforgeeksisforgeeks"; 
    n = len(s); 
  
    print(countSubStr(s, n)); 
      
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
      
public class GFG 
{
   
    // Function to return the count
    // of required substrings
    static int countSubStr(String s, int n)
    {
        int c1 = 0, c2 = 0;
   
        // For every index of the string
        for (int i = 0; i < n; i++)
        {
   
            // If the substring starting at
            // the current index is "geeks"
            if (i < n - 5 && 
                "geeks".Equals(s.Substring(i, 5))) 
            {
                c1++;
            }
   
            // If the substring is "for"
            if (i < n - 3 && 
                "for".Equals(s.Substring(i, 3))) 
            {
                c2 = c2 + c1;
            }
        }
        return c2;
    }
   
    // Driver code
    public static void Main(String[] args)
    {
        String s = "geeksforgeeksisforgeeks";
        int n = s.Length;
        Console.WriteLine(countSubStr(s, n));
    }
}
  
// This code is contributed by 29AjayKumar

输出：

3