通过交换位来更改两个数字的XOR的方法的数量

给定两个二进制字符串s1和s2。它们的XOR为X，任务是找到交换字符串s1中两位位置的方式的数量，以使新s1和s2之间形成的XOR与X不同。
例子：

Input: s1 = “01011”, s2 = “11001” 
Output: 4 
swap bits of index(1-based) (1, 4), (2, 3), (3, 4), or (3, 5) such that XOR value is changed. 
Input: s1 = “011000”, s2 = “010011” 
Output: 6

方法：

1. 计算s1中1和0的数量。
2. 遍历字符串s1，并检查以下两种情况：
   • s1 [i]和s2 [i]中的0和0 ，用0代替1将改变XOR值。
   • 在S1中1和0 [i]和S2 [I]中，与0代替1将改变异或值。
3. 对于第一种情况，替换方式的数量将为已使用的1的数量。
4. 对于第二种情况，替换方式的数量将是零的数量-已经使用的0。
5. 两种情况下方式总数的总和就是答案。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
// Function that returns the number of
// bit swaps such that xor is different
int countWays(string s1, string s2)
{
    int c1 = 0, c0 = 0;
    int n = s1.length();
 
    // traverse and count 1's and 0's
    for (int i = 0; i < n; i++) {
        if (s1[i] == '1')
            c1++;
        else
            c0++;
    }
    int used1 = 0, used0 = 0;
    int ways = 0;
 
    // traverse in the string
    for (int i = 0; i < n; i++) {
 
        // if both positions are 0
        if (s1[i] == '0' and s2[i] == '0') {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR, we have to
        // replace 1 by 0
        else if (s1[i] == '1' and s2[i] == '0') {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
int main()
{
    string s1 = "01011";
    string s2 = "11001";
 
    cout << countWays(s1, s2);
    return 0;
}

Java

// Java Program to find Number of
// ways to change the XOR of two
// numbers by swapping the bits
class GFG
{
// Function that returns the
// number of bit swaps such
// that xor is different
static int countWays(String s1,
                     String s2)
{
    int c1 = 0, c0 = 0;
    int n = s1.length();
 
    // traverse and count 1's and 0's
    for (int i = 0; i < n; i++)
    {
        if (s1.charAt(i) == '1')
            c1++;
        else
            c0++;
    }
    int used1 = 0, used0 = 0;
    int ways = 0;
 
    // traverse in the String
    for (int i = 0; i < n; i++)
    {
 
        // if both positions are 0
        if (s1.charAt(i) == '0' &&
            s2.charAt(i) == '0')
        {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of
            // ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR,
        // we have to replace 1 by 0
        else if (s1.charAt(i) == '1' &&
                 s2.charAt(i) == '0')
        {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of
            // 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
public static void main(String[] args)
{
    String s1 = "01011";
    String s2 = "11001";
 
    System.out.println(countWays(s1, s2));
}
}
 
// This code is contributed
// by Arnab Kundu

Python3

# Function that returns the number of
# bit swaps such that xor is different
def countWays(s1, s2):
 
    c1 = 0
    c0 = 0
    n = len(s1)
 
    # traverse and count 1's and 0's
    for i in range(0,n) :
        if (s1[i] == '1'):
            c1+=1
        else:
            c0+=1
     
    used1 = 0
    used0 = 0
    ways = 0
 
    # traverse in the string
    for i in range(0,n) :
 
        # if both positions are 0
        if (s1[i] == '0' and s2[i] == '0') :
 
            # add the number of ones as
            # it will change the XOR
            ways += c1
 
            # subtract the number of ones already used
            ways -= used1
 
            # zeros have been used
            used0+=1
         
 
        # when 1 and 0, to change XOR, we have to
        # replace 1 by 0
        elif (s1[i] == '1' and s2[i] == '0') :
 
            # add number of 0's
            ways += c0
 
            # subtract number of 0's already used
            ways -= used0
 
            # count 1's used
            used1+=1
 
    # return the answer
    return ways
 
# Driver Code
if __name__=='__main__':
    s1 = "01011"
    s2 = "11001"
    print(countWays(s1, s2))
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# Program to find Number of
// ways to change the XOR of two
// numbers by swapping the bits
using System;
 
class GFG
{
// Function that returns the
// number of bit swaps such
// that xor is different
static int countWays(String s1,
                     String s2)
{
    int c1 = 0, c0 = 0;
    int n = s1.Length;
 
    // traverse and count 1's and 0's
    for (int i = 0; i < n; i++)
    {
        if (s1[i] == '1')
            c1++;
        else
            c0++;
    }
    int used1 = 0, used0 = 0;
    int ways = 0;
 
    // traverse in the String
    for (int i = 0; i < n; i++)
    {
 
        // if both positions are 0
        if (s1[i] == '0' &&
            s2[i] == '0')
        {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of
            // ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR,
        // we have to replace 1 by 0
        else if (s1[i] == '1' &&
                 s2[i] == '0')
        {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of
            // 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
public static void Main(String[] args)
{
    String s1 = "01011";
    String s2 = "11001";
 
    Console.WriteLine(countWays(s1, s2));
}
}
 
// This code is contributed
// by Subhadeep Gupta

Javascript

 
// Function that returns the number of
// bit swaps such that xor is different
function countWays(s1, s2)
{
    let c1 = 0, c0 = 0;
    let n = s1.length;
 
    // traverse and count 1's and 0's
    for (let i = 0; i < n; i++) {
        if (s1[i] == '1')
            c1++;
        else
            c0++;
    }
    let used1 = 0, used0 = 0;
    let ways = 0;
 
    // traverse in the string
    for (let i = 0; i < n; i++) {
 
        // if both positions are 0
        if (s1[i] == '0' && s2[i] == '0') {
 
            // add the number of ones as
            // it will change the XOR
            ways += c1;
 
            // subtract the number of ones already used
            ways -= used1;
 
            // zeros have been used
            used0++;
        }
 
        // when 1 and 0, to change XOR, we have to
        // replace 1 by 0
        else if (s1[i] == '1' && s2[i] == '0') {
 
            // add number of 0's
            ways += c0;
 
            // subtract number of 0's already used
            ways -= used0;
 
            // count 1's used
            used1++;
        }
    }
 
    // return the answer
    return ways;
}
 
// Driver Code
    let s1 = "01011";
    let s2 = "11001";
 
    document.write(countWays(s1, s2));
 

输出：

4

时间复杂度： O(N)

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