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：下面给出的表格解释了XOR和OR操作的所有可能性。

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，则可以进行所有转换，可以在表中看到

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)

Javascript

// Javascript program to check if string1 can be
// converted to string2 using XOR and OR operations
 
// function to check if conversion is possible or not
function solve(s1, s2)
{
    let flag1 = 0, flag2 = 0;
 
    // if lengths are different
    if (s1.length != s2.length)
        return false;
     
    let l = s1.length;
 
    // iterate to check if both strings have 1
    for (let 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
    let s1 = "100101";
    let s2 = "100000";
 
    if (solve(s1, s2))
        document.write("Yes");
    else
        document.write("No");
 

输出：

Yes

时间复杂度： O(n)，其中n是输入字符串的长度。

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