在第二个字符串上执行交换后，找到两个字符串之间的最长公共前缀

给定两个字符串$a$和$b$ 。在对字符串执行零次或多次操作后，找到它们之间最长的公共前缀$b$ 。在每个操作中，您可以交换任意两个字母。

例子：

Input : a = "here", b = "there"
Output : 4
The 2nd string can be made "heret" by just 
swapping characters and thus the longest
prefix is of length 4.

Input : a = "you", b = "me"
Output : 0

鉴于我们只允许在字符串执行交换$b$并且前缀的长度应最大化。所以这个想法是遍历字符串$a$并检查字符串中当前字符的频率$a$与字符串的相同或更少$b$ 。如果是的话再往前走的字符串$a$以其他方式分解，并打印到其中字符在字符串匹配字符串的部分的长度$b$ 。

下面是上述方法的实现：

C++

// C++ program to find the longest
// common prefix between two strings 
// after performing swaps on the second string
#include 
using namespace std;
  
   
void LengthLCP(string x, string y)
{
      
  
    int fr[26]={0};
       
    int a = x.length(); // length of x
    int b = y.length(); // length of y
       
    for (int i=0 ;i 0){
            c += 1;
            fr[x[i] - 97] -= 1;
        }
        else
            break;
    }
    cout<<(c)<

---

Java

// Java program to find the longest
// common prefix between two strings 
// after performing swaps on the second string
  
public class GFG {
  
    static void LengthLCP(String x, String y)
    {
          
  
        int fr[]=new int [26];
           
        int a = x.length(); // length of x
        int b = y.length(); // length of y
           
        for (int i=0 ;i 0){
                c += 1;
                fr[x.charAt(i) - 97] -= 1;
            }
            else
                break;
        }
        System.out.println((c)) ;
    }
  
  
    public static void main(String args[])
    {
        String x="here", y =  "there";
           
        LengthLCP(x, y);
  
  
    }
    // This code is contributed by ANKITRAI1
}

---

Python3

# Python program to find the longest
# common prefix between two strings 
# after performing swaps on the second string
  
def LengthLCP(x, y):
    fr = [0] * 26
      
    a = len(x) # length of x
    b = len(y) # length of y
      
    for i in range(b):
        # creating frequency array of
        # characters of y
        fr[ord(y[i]) - 97] += 1
      
    # storing the length of 
    # longest common prefix
    c = 0 
      
    for i in range(a):
        # checking if the frequency of the character at
        # position i in x in b is greater than zero or not
        # if zero we increase the prefix count by 1
        if (fr[ord(x[i]) - 97] > 0):
            c += 1
            fr[ord(x[i]) - 97] -= 1
        else:
            break
    print(c)
  
# Driver Code
  
x, y = "here", "there"
  
LengthLCP(x, y)

C#

// C# program to find the longest 
// common prefix between two strings 
// after performing swaps on the 
// second string
using System;
  
class GFG 
{ 
  
static void LengthLCP(String x, String y) 
{ 
    int []fr = new int [26]; 
      
    int a = x.Length; // length of x 
    int b = y.Length; // length of y 
      
    for (int i = 0 ; i < b; i++) 
    { 
        // creating frequency array 
        // of characters of y 
        fr[y[i] - 97] += 1; 
    } 
      
    // storing the length of 
    // longest common prefix 
    int c = 0; 
      
    for (int i = 0 ; i < a; i++) 
    { 
        // checking if the frequency of 
        // the character at position i
        // in x in b is greater than zero 
        // or not if zero we increase the 
        // prefix count by 1 
        if (fr[x[i] - 97] > 0)
        { 
            c += 1; 
            fr[x[i] - 97] -= 1; 
        } 
        else
            break; 
    } 
    Console.Write((c)) ; 
} 
  
// Driver Code
public static void Main() 
{ 
    String x = "here", y = "there"; 
      
    LengthLCP(x, y); 
} 
} 
  
// This code is contributed by 29AjayKumar

PHP

 0)
        {
            $c += 1;
            $fr[ord($x[$i]) - 97] -= 1;
        }
        else
            break;
    }
    echo $c;
}
  
// Driver Code
$x="here";
$y = "there";
  
LengthLCP($x, $y);
  
return 0;
  
// This code is contributed by ChitraNayal
?>

输出：

4