检查是否可以通过交换给定类型的相邻字符来将字符串转换为另一个字符串

给定两个大小为N 的字符串str1和str2仅由三个字符A 、 B和C 组成，任务是检查字符串str1是否可以使用以下操作更改为str2 ：

• 用“CB”替换一次出现的“BC”，即交换相邻的“B”和“C”。
• 用“AC”替换一次出现的“CA”，即交换相邻的“C”和“A”。

如果我们可以转换字符串，则打印“是” ，否则打印“否” 。
例子：

Input: str1 = “BCCABCBCA”, str2 = “CBACCBBAC” 
Output: Yes 
Explanation: 
Transform the strings using following these steps: 
BCCABCBCA -> CBCABCBCA -> CBACBCBCA -> CBACCBBCA -> CBACCBBAC.
Input: str1 = “BAC”, str2 = “CAB” 
Output: False

朴素的方法：这个想法是通过执行给定的操作从字符串str1的开头递归地生成所有可能的字符串，并将其存储在一组字符串。然后检查集合中的任何字符串是否等于字符串str2 。如果在集合中找到字符串str2则打印“是”，否则打印“否” 。

时间复杂度： O(2 ^N )
辅助空间： O(1)
有效的方法：这个想法是同时遍历两个字符串并检查是否可以将字符串str1转换为str2直到特定索引。以下是步骤：

1. 检查字符串str1和str2 中的序列 ‘A’ 和 ‘B’ ，如果相同，则进行第二步。否则，打印“否”，因为无法进行所需的转换。
2. str1中’A’ 的索引应该大于等于str2 中相应’A’ 的索引，因为“CA”只能转换为“AC”。
3. 同样， str1字符串中 ‘B’ 的索引应该小于或等于str2中 ‘B’ 的对应索引，因为“BC”只能转换为“CB”。
4. 如果不满足以上两个条件，则打印“否” 。否则，打印“是” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if it is possible
// to transform start to end
bool canTransform(string str1,
                string str2)
{
    string s1 = "";
    string s2 = "";
 
    // Check the sequence of A, B in
    // both strings str1 and str2
    for (char c : str1) {
        if (c != 'C') {
            s1 += c;
        }
    }
 
    for (char c : str2) {
        if (c != 'C') {
            s2 += c;
        }
    }
 
    // If both the strings
    // are not equal
    if (s1 != s2)
        return false;
 
    int i = 0;
    int j = 0;
    int n = str1.length();
 
    // Traverse the strings
    while (i < n and j < n) {
        if (str1[i] == 'C') {
            i++;
        }
 
        else if (str2[j] == 'C') {
            j++;
        }
 
        // Check for indexes of A and B
        else {
            if ((str1[i] == 'A'
                and i < j)
                or (str1[i] == 'B'
                    and i > j)) {
                return false;
            }
            i++;
            j++;
        }
    }
 
    return true;
}
 
// Driver Code
int main()
{
    string str1 = "BCCABCBCA";
    string str2 = "CBACCBBAC";
 
    // Function Call
    if (canTransform(str1, str2)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
    // Function to check if it is possible
    // to transform start to end
    static boolean canTransform(String str1, String str2)
    {
        String s1 = "";
        String s2 = "";
 
        // Check the sequence of A, B in
        // both Strings str1 and str2
        for (char c : str1.toCharArray())
        {
            if (c != 'C')
            {
                s1 += c;
            }
        }
 
        for (char c : str2.toCharArray())
        {
            if (c != 'C')
            {
                s2 += c;
            }
        }
 
        // If both the Strings
        // are not equal
        if (!s1.equals(s2))
            return false;
 
        int i = 0;
        int j = 0;
        int n = str1.length();
 
        // Traverse the Strings
        while (i < n && j < n)
        {
            if (str1.charAt(i) == 'C')
            {
                i++;
            }
            else if (str2.charAt(j) == 'C')
            {
                j++;
            }
 
            // Check for indexes of A and B
            else
            {
                if ((str1.charAt(i) == 'A' && i < j) ||
                    (str1.charAt(i) == 'B' && i > j))
                {
                    return false;
                }
                i++;
                j++;
            }
        }
        return true;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String str1 = "BCCABCBCA";
        String str2 = "CBACCBBAC";
 
        // Function Call
        if (canTransform(str1, str2))
        {
            System.out.print("Yes");
        }
        else
        {
            System.out.print("No");
        }
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
 
# Function to check if it is possible
# to transform start to end
def canTransform(str1, str2):
     
    s1 = ""
    s2 = ""
 
    # Check the sequence of A, B in
    # both strings str1 and str2
    for c in str1:
        if (c != 'C'):
            s1 += c
 
    for c in str2:
        if (c != 'C'):
            s2 += c
 
    # If both the strings
    # are not equal
    if (s1 != s2):
        return False
 
    i = 0
    j = 0
    n = len(str1)
 
    # Traverse the strings
    while (i < n and j < n):
        if (str1[i] == 'C'):
            i += 1
 
        elif (str2[j] == 'C'):
            j += 1
 
        # Check for indexes of A and B
        else:
            if ((str1[i] == 'A' and i < j) or
                (str1[i] == 'B' and i > j)):
                return False
                 
            i += 1
            j += 1
 
    return True
 
# Driver Code
if __name__ == '__main__':
     
    str1 = "BCCABCBCA"
    str2 = "CBACCBBAC"
 
    # Function call
    if (canTransform(str1, str2)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if it is possible
// to transform start to end
static bool canTransform(string str1, string str2)
{
    string s1 = "";
    string s2 = "";
 
    // Check the sequence of A, B in
    // both Strings str1 and str2
    foreach(char c in str1.ToCharArray())
    {
        if (c != 'C')
        {
            s1 += c;
        }
    }
 
    foreach(char c in str2.ToCharArray())
    {
        if (c != 'C')
        {
            s2 += c;
        }
    }
 
    // If both the Strings
    // are not equal
    if (s1 != s2)
        return false;
 
    int i = 0;
    int j = 0;
    int n = str1.Length;
 
    // Traverse the Strings
    while (i < n && j < n)
    {
        if (str1[i] == 'C')
        {
            i++;
        }
        else if (str2[j] == 'C')
        {
            j++;
        }
 
        // Check for indexes of A and B
        else
        {
            if ((str1[i] == 'A' && i < j) ||
                (str1[i] == 'B' && i > j))
            {
                return false;
            }
            i++;
            j++;
        }
    }
    return true;
}
 
// Driver Code
public static void Main(string[] args)
{
    string str1 = "BCCABCBCA";
    string str2 = "CBACCBBAC";
 
    // Function call
    if (canTransform(str1, str2))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by rutvik_56

Javascript

      // JavaScript program for the above approach
      // Function to check if it is possible
      // to transform start to end
      function canTransform(str1, str2) {
        var s1 = "";
        var s2 = "";
 
        // Check the sequence of A, B in
        // both Strings str1 and str2
        for (const c of str1) {
          if (c !== "C") {
            s1 += c;
          }
        }
 
        for (const c of str2) {
          if (c !== "C") {
            s2 += c;
          }
        }
 
        // If both the Strings
        // are not equal
        if (s1 !== s2) return false;
 
        var i = 0;
        var j = 0;
        var n = str1.length;
 
        // Traverse the Strings
        while (i < n && j < n) {
          if (str1[i] === "C") {
            i++;
          } else if (str2[j] === "C") {
            j++;
          }
 
          // Check for indexes of A and B
          else {
            if ((str1[i] === "A" && i < j) || (str1[i] === "B" && i > j)) {
              return false;
            }
            i++;
            j++;
          }
        }
        return true;
      }
 
      // Driver Code
      var str1 = "BCCABCBCA";
      var str2 = "CBACCBBAC";
 
      // Function call
      if (canTransform(str1.split(""), str2.split(""))) {
        document.write("Yes");
      } else {
        document.write("No");
      }
    

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)

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