在对第二个字符串执行交换后找到两个字符串之间最长的公共前缀

给定两个字符串$a$和$b$ .对字符串执行零个或多个操作后，找出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$b$并且前缀的长度应该最大化。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，那么在字符串向前否则突破并打印高达字符所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
?>

Javascript

 
// JavaScript program to find the long
// common prefix between two strings
// after performing swaps on the second string
 
    function LengthLCP(x, y)
    {
           
   
        let fr = Array(26).fill(0);
            
        let a = x.length; // length of x
        let b = y.length; // length of y
            
        for (let i=0 ;i<b ; i++)
        {
            // creating frequency array of
            // characters of y
            fr[y[i].charCodeAt() - 97] += 1;
        }
        // storing the length of
        // longest common prefix
        let c = 0;
            
        for (let 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].charCodeAt() - 97] > 0){
                c += 1;
                fr[x[i].charCodeAt() - 97] -= 1;
            }
            else
                break;
        }
        document.write((c)) ;
    }
 
 
// driver code
 
    let x="here", y =  "there";
            
        LengthLCP(x, y);
   

输出：

4

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