计数的字符从字符串以获得另一字符串的开始或结束错开

给定两个字符串A和B ，其中字符串A是字符串B的字谜。在一个操作中，删除字符串A的第一个或最后一个字符，然后在A中的任何位置插入。任务是找到将字符串A转换为字符串B所需执行的此类操作的最少数量。

例子：

Input: A = “edacb”, B = “abcde”
Output: 3
Explanation:
Perform the given operations as follows:
Step1: Take last character ‘b’ and insert it between ‘a’ and ‘c’ ( “edacb” becomes “edabc” )
Step2: Take first character ‘e’ insert it to last ( “edabc” becomes “dabce” )
Step3: Take first character ‘d’ and insert it between ‘c’ and ‘e’ ( “dabce” becomes “abcde” )
Therefore, the count of the operations is 3.

Input: A = “abcd”, B = “abdc”
Output: 1
Explanation:
Perform the given operations as follows:
Step1: Take last character ‘d’ and insert it between ‘b’ and ‘c’ ( “abcd” becomes “abdc” )
Threfore, the count of the operations is 1.

方法：这个想法是找到字符串A的最长子字符串，它也是字符串B的子序列，并从给定的字符串长度中减去该值以获得所需的最小操作数。

下面是上述方法的实现：

C++14

// C++14 program for the above approach
#include 
using namespace std;
 
// Function to find the minimum cost
// to convert string A to string B
int minCost(string A, string B)
{
     
    // Length of string
    int n = A.size();
  
    int i = 0;
  
    // Initialize maxlen as 0
    int maxlen = 0;
  
    // Traverse the string A
    while (i < n)
    {
         
        // Stores the length of
        // substrings of string A
        int length = 0;
  
        // Traversing string B for
        // each character of A
        for(int j = 0; j < n; ++j)
        {
             
            // Shift i pointer towards
            // right and increment length,
            // if A[i] equals B[j]
            if (A[i] == B[j])
            {
                ++i;
                ++length;
  
                // If traverse till end
                if (i == n)
                break;
            }
        }
         
        // Update maxlen
        maxlen = max(maxlen,
                     length);
    }
     
    // Return minimum cost
    return n - maxlen;
}
  
// Driver Code
int main()
{
     
    // Given two strings A and B
    string A = "edacb";
    string B = "abcde";
  
    // Function call
    cout << minCost(A, B) << endl;
}
 
// This code is contributed by sanjoy_62

Java

// Java Program for the above approach
 
import java.util.*;
import java.lang.*;
 
class GFG {
 
    // Function to find the minimum cost
    // to convert string A to string B
    static int minCost(String A, String B)
    {
        // Length of string
        int n = A.length();
 
        int i = 0;
 
        // Initialize maxlen as 0
        int maxlen = 0;
 
        // Traverse the string A
        while (i < n) {
 
            // Stores the length of
            // substrings of string A
            int length = 0;
 
            // Traversing string B for
            // each character of A
            for (int j = 0; j < n; ++j) {
 
                // Shift i pointer towards
                // right and increment length,
                // if A[i] equals B[j]
                if (A.charAt(i) == B.charAt(j)) {
 
                    ++i;
                    ++length;
 
                    // If traverse till end
                    if (i == n)
                        break;
                }
            }
 
            // Update maxlen
            maxlen = Math.max(maxlen,
                              length);
        }
 
        // Return minimum cost
        return n - maxlen;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given two strings A and B
        String A = "edacb";
        String B = "abcde";
 
        // Function call
        System.out.println(minCost(A, B));
    }
}

Python3

# Python3 Program for
# the above approach
# Function to find the
# minimum cost to convert
# string A to string B
def minCost(A, B):
   
    # Length of string
    n = len(A);
 
    i = 0;
 
    # Initialize maxlen as 0
    maxlen = 0;
 
    # Traverse the string A
    while (i < n):
 
        # Stores the length of
        # substrings of string A
        length = 0;
 
        # Traversing string B for
        # each character of A
        for j in range(0, n):
 
            # Shift i pointer towards
            # right and increment length,
            # if A[i] equals B[j]
            if (A[i] == B[j]):
 
                i+= 1
                length+=1;
 
                # If traverse till end
                if (i == n):
                    break;
 
        # Update maxlen
        maxlen = max(maxlen, length);
 
    # Return minimum cost
    return n - maxlen;
 
# Driver Code
if __name__ == '__main__':
   
    # Given two strings A and B
    A = "edacb";
    B = "abcde";
 
    # Function call
    print(minCost(A, B));
 
# This code is contributed by Rajput-Ji

C#

// C# program for the above approach 
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the minimum cost
// to convert string A to string B
static int minCost(string A, string B)
{
     
    // Length of string
    int n = A.Length;
 
    int i = 0;
 
    // Initialize maxlen as 0
    int maxlen = 0;
 
    // Traverse the string A
    while (i < n)
    {
         
        // Stores the length of
        // substrings of string A
        int length = 0;
 
        // Traversing string B for
        // each character of A
        for(int j = 0; j < n; ++j)
        {
             
            // Shift i pointer towards
            // right and increment length,
            // if A[i] equals B[j]
            if (A[i] == B[j])
            {
                ++i;
                ++length;
 
                // If traverse till end
                if (i == n)
                    break;
            }
        }
 
        // Update maxlen
        maxlen = Math.Max(maxlen,
                          length);
    }
 
    // Return minimum cost
    return n - maxlen;
}
 
// Driver Code
public static void Main()
{
     
    // Given two strings A and B
    string A = "edacb";
    string B = "abcde";
 
    // Function call
    Console.WriteLine(minCost(A, B));
}
}
 
// This code is contributed by sanjoy_62

输出：

3

时间复杂度： O(N ² )，其中N是字符串的长度
辅助空间： O(N)