使用XOR和OR进行字符串转换

给定两个二进制字符串。任务是检查是否可以通过多次执行给定的操作将字符串s1转换为字符串s2。

• 选择字符串s1中的任意两个相邻字符，然后用a ^ b替换其中一个，用a替换另一个$\oplus$ b(a OR b)。

例子：

Input: S1 = “11”, S2 = “10”
Output: YES
Select two adjacent characters and replace s2[0] by s1[0]$\oplus$s1[1] and
change s2[1] by s1[0]^s1[1]

Input: S1 = “000”, S2 = “101”
Output: NO

方法：下表给出了解释XOR和OR操作的所有可能性的表格。

X	Y	X^Y	X$\oplus$Y
0	0	0	0
0	1	1	1
1	0	1	1
1	1	0	1

如果两个字符串都仅由0组成并且长度相同，则可以进行转换，因为两个相邻的零将仅产生零，而与对其进行的操作无关。如果两个字符串都为1，请按照以下步骤检查String1是否可以转换为String2。

• 检查长度是否相等
• 检查两个字符串是否都至少为1，因为如果两个字符串都至少为1(在表中可以看到)，则可以进行所有转换

如果以上两个条件都成立，则可以将String1转换为String2。

下面是上述方法的实现：

C++

// C++ program to check if string1 can be
// converted to string2 using XOR and OR operations
#include 
using namespace std;
  
// function to check if conversion is possible or not
bool solve(string s1, string s2)
{
    bool flag1 = 0, flag2 = 0;
  
    // if lengths are different
    if (s1.length() != s2.length()) 
        return false;
      
    int l = s1.length();
  
    // iterate to check if both strings have 1
    for (int i = 0; i < l; i++) {
  
        // to check if there is
        // even one 1 in string s1
        if (s1[i] == '1')
            flag1 = 1;
  
        // to check if there is even
        // one 1 in string s2
        if (s2[i] == '1')
            flag2 = 1;
  
        if (flag1 && flag2)
            return true;
    }
     //if both strings have only '0'
     if(!flag1&&!flag2)
            return true;
    // if both string do not have a '1'.
    return false;
}
  
// Driver code
int main()
{
    string s1 = "100101";
    string s2 = "100000";
  
    if (solve(s1, s2))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to check if
// string1 can be converted
// to string2 using XOR and
// OR operations
import java.io.*;
import java.util.*;
  
class GFG
{
      
// function to check if 
// conversion is possible
// or not
static boolean solve(String s1,
                     String s2)
{
    boolean flag1 = false, 
            flag2 = false;
  
    // if lengths are different
    if (s1.length() != s2.length()) 
        return false;
      
    int l = s1.length();
  
    // iterate to check if
    // both strings have 1
    for (int i = 0; i < l; i++) 
    {
  
        // to check if there is
        // even one 1 in string s1
        if (s1.charAt(i) == '1')
            flag1 = true;
  
        // to check if there is even
        // one 1 in string s2
        if (s2.charAt(i) == '1')
            flag2 = true;
  
        if (flag1 == true && 
            flag2 == true)
            return true;
    }
     //if both strings have only '0'
     if(!flag1&&!flag2)
            return true;
    // if both string do 
    // not have a '1'.
    return false;
}
  
// Driver code
public static void main(String args[])
{
    String s1 = "100101";
    String s2 = "100000";
  
    if (solve(s1, s2) == true)
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

Python3

# Python3 program to check
# if string1 can be converted 
# to string2 using XOR and 
# OR operations
  
# function to check if 
# conversion is possible or not
def solve(s1, s2):
    flag1 = 0
    flag2 = 0
  
# if lengths are different
    if (len(s1) != len(s2)): 
        return False
      
    l = len(s1)
  
# iterate to check if
# both strings have 1
    for i in range (0, l): 
  
    # to check if there is
    # even one 1 in string s1
        if (s1[i] == '1'):
            flag1 = 1;
  
    # to check if there is even
    # one 1 in string s2
        if (s2[i] == '1'):
            flag2 = 1
  
    # if both string 
    # do not have a '1'.
        if (flag1 & flag2):
            return True
  
        if(!flag1 & !flag2):
            return True
    return False
  
# Driver code
s1 = "100101"
s2 = "100000"
  
if solve(s1, s2):
    print( "Yes")
else:
    print("No")
  
# This code is contributed 
# by Shivi_Aggarwal

C#

// C# program to check if
// string1 can be converted
// to string2 using XOR and
// OR operations
using System;
  
class GFG
{
      
// function to check if 
// conversion is possible
// or not
static bool solve(String s1,
                  String s2)
{
    bool flag1 = false, 
         flag2 = false;
  
    // if lengths are different
    if (s1.Length != s2.Length)
        return false;
      
    int l = s1.Length;
  
    // iterate to check if
    // both strings have 1
    for (int i = 0; i < l; i++) 
    {
  
        // to check if there is
        // even one 1 in string s1
        if (s1[i] == '1')
            flag1 = true;
  
        // to check if there is even
        // one 1 in string s2
        if (s2[i] == '1')
            flag2 = true;
  
        if (flag1 == true && 
            flag2 == true)
            return true;
    }
  
     //if both strings have only '0'
     if(!flag1&&!flag2)
            return true;
    // if both string do 
    // not have a '1'.
    return false;
}
  
// Driver code
public static void Main()
{
    String s1 = "100101";
    String s2 = "100000";
  
    if (solve(s1, s2) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
} 
  
// This code is contributed
// by Akanksha Rai(Abby_akku)

输出：

Yes

时间复杂度： O(n)，其中n是输入字符串的长度。