查找其字符可以重新排列以形成给定单词的子字符串数

给定一个字符串str ，任务是查找所有长度为4的子字符串的计数，这些子字符串的字符可以重新排列以形成单词“ clap” 。

例子：

Input: str = “clapc”
Output: 2
“clap” and “lapc” are the required sub-strings

Input: str = “abcd”
Output: 0

方法：对于长度为四的每个子字符串，计算单词“ clap”中字符的出现次数。如果每个字符都在子字符串中恰好出现一次，则增加count 。最后打印计数。

下面是上述方法的实现：

C++

// CPP implementation of the approach
#include
using namespace std;
  
// Function to return the count of
// required occurrence
int countOcc(string s)
{
  
    // To store the count of occurrences
    int cnt = 0;
  
    // Check first four characters from ith position
    for (int i = 0; i < s.length() - 3; i++) 
    {
  
        // Variables for counting the required characters
        int c = 0, l = 0, a = 0, p = 0;
  
        // Check the four contiguous characters which
        // can be reordered to form 'clap'
        for (int j = i; j < i + 4; j++) 
        {
            switch (s[j]) 
            {
                case 'c':
                    c++;
                    break;
                case 'l':
                    l++;
                    break;
                case 'a':
                    a++;
                    break;
                case 'p':
                    p++;
                    break;
            }
        }
  
        // If all four contiguous characters are present
        // then increment cnt variable
        if (c == 1 && l == 1 && a == 1 && p == 1)
            cnt++;
    }
  
    return cnt;
}
  
// Driver code
int main()
{
    string s = "clapc";
    transform(s.begin(), s.end(), s.begin(), ::tolower);
    cout << (countOcc(s));
}
  
// This code is contributed by
// Surendra_Gangwar

Java

// Java implementation of the approach
class GFG {
  
    // Function to return the count of
    // required occurrence
    static int countOcc(String s)
    {
  
        // To store the count of occurrences
        int cnt = 0;
  
        // Check first four characters from ith position
        for (int i = 0; i < s.length() - 3; i++) {
  
            // Variables for counting the required characters
            int c = 0, l = 0, a = 0, p = 0;
  
            // Check the four contiguous characters which
            // can be reordered to form 'clap'
            for (int j = i; j < i + 4; j++) {
                switch (s.charAt(j)) {
                case 'c':
                    c++;
                    break;
                case 'l':
                    l++;
                    break;
                case 'a':
                    a++;
                    break;
                case 'p':
                    p++;
                    break;
                }
            }
  
            // If all four contiguous characters are present
            // then increment cnt variable
            if (c == 1 && l == 1 && a == 1 && p == 1)
                cnt++;
        }
  
        return cnt;
    }
  
    // Driver code
    public static void main(String args[])
    {
        String s = "clapc";
        System.out.print(countOcc(s.toLowerCase()));
    }
}

Python3

# Python3 implementation of the approach
  
# Function to return the count of
# required occurrence
def countOcc(s):
  
    # To store the count of occurrences
    cnt = 0
  
    # Check first four characters from ith position
    for i in range(0, len(s) - 3): 
  
        # Variables for counting the required characters
        c, l, a, p = 0, 0, 0, 0
  
        # Check the four contiguous characters
        # which can be reordered to form 'clap'
        for j in range(i, i + 4): 
              
            if s[j] == 'c':
                c += 1
          
            elif s[j] == 'l':
                l += 1
                  
            elif s[j] == 'a':
                a += 1
                  
            elif s[j] == 'p':
                p += 1
  
        # If all four contiguous characters are
        # present then increment cnt variable
        if c == 1 and l == 1 and a == 1 and p == 1:
            cnt += 1
          
    return cnt
      
# Driver code
if __name__ == "__main__":
      
    s = "clapc"
    print(countOcc(s.lower()))
      
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to return the count of
// required occurrence
static int countOcc(string s)
{
  
    // To store the count of occurrences
    int cnt = 0;
  
    // Check first four characters 
    // from ith position
    for (int i = 0; i < s.Length - 3; i++) 
    {
  
        // Variables for counting the 
        // required characters
        int c = 0, l = 0, a = 0, p = 0;
  
        // Check the four contiguous characters 
        // which can be reordered to form 'clap'
        for (int j = i; j < i + 4; j++) 
        {
            switch (s[j]) 
            {
                case 'c':
                    c++;
                    break;
                case 'l':
                    l++;
                    break;
                case 'a':
                    a++;
                    break;
                case 'p':
                    p++;
                    break;
            }
        }
  
        // If all four contiguous characters are
        // present then increment cnt variable
        if (c == 1 && l == 1 && a == 1 && p == 1)
            cnt++;
    }
  
    return cnt;
}
  
// Driver code
public static void Main()
{
    string s = "clapc";
    Console.Write(countOcc(s.ToLower()));
}
}
  
// This code is contributed by Akanksha Rai

输出：

2