将一个二进制字符串转换为另一个二进制字符串所需的最少交换

给定两个长度相等的二进制字符串M和N ，任务是找到将字符串N转换为M所需的最小操作数(交换)。

例子：

Input: str1 = "1101", str2 = "1110"
Output: 1
Swap last and second last element in the binary string, 
so that it become 1101

Input: str1 = "1110000", str2 = "0001101"
Output: 3

方法：初始化计数器，并在M上进行迭代，以便在两个二进制字符串中都发现任何不相等的元素时，增加计数器。最后，如果计数器为偶数，则输出结果/ 2，因为对于一次交换，两个元素是不相同的。
假设S1 =“ 10”且S2 =“ 01” ，那么两对不相同，计数= 2，并且当计数为偶数时，交换数为count / 2，即1。偶数确定存在交换元素的机会。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Method to count swaps
void minSwaps(string str1, string str2)
{
    // Initialize the count
    int count = 0;
  
    // Iterate the loop with str1 length
    for (int i = 0; i < str1.length(); i++) {
  
        // If any non-equal elements are found
        // increment the counter
        if (str1[i] != str2[i])
            count++;
    }
  
    // If counter is even print the swap
    if (count % 2 == 0)
        cout << count / 2;
    else
        cout << "Not Possible";
}
  
// Driver code
int main()
{
    // Take two input
    string binaryString1 = "1110000";
    string binaryString2 = "0001101";
  
    // Call the method
    minSwaps(binaryString1, binaryString2);
  
    return 0;
}

Java

// Java Program to count minimum number of swap
// required to make string N to M
public class GFG {
  
    // Method to count swaps
    static void minSwaps(String str1, String str2)
    {
        // Initialize the count
        int count = 0;
  
        // Iterate the loop with str1 length
        for (int i = 0; i < str1.length(); i++) {
  
            // If any non-equal elements are found
            // increment the counter
            if (str1.charAt(i) != str2.charAt(i))
                count++;
        }
  
        // If counter is even print the swap
        if (count % 2 == 0)
            System.out.println(count / 2);
        else
            System.out.println("Not Possible");
    }
  
    // Driver Code
    public static void main(String args[])
    {
        // Take two input
        String binaryString1 = "1110000";
        String binaryString2 = "0001101";
  
        // Call the method
        minSwaps(binaryString1, binaryString2);
    }
}

Python 3

# Python3 implementation of 
# the above approach 
  
# function to count swaps
def minSwaps(str1, str2) :
  
    # Initialize the count
    count = 0
  
    # Iterate the loop with 
    # length of str1
    for i in range(len(str1)) :
  
        # If any non-equal elements are 
        # found increment the counter 
        if str1[i] != str2[i] :
            count += 1
  
    # If counter is even print 
    # the swap 
    if count % 2 == 0 :
        print(count // 2)
    else :
        print("Not Possible")
  
  
# Driver code
if __name__ == "__main__" :
  
    # Take two input
    binaryString1 = "1110000"
    binaryString2 = "0001101"
  
    # Call the function
    minSwaps( binaryString1, binaryString2)
  
# This code is contributed by ANKITRAI1

C#

// C# Program to count minimum number of swap
// required to make string N to M
using System;
class GFG
{
  
// Method to count swaps
static void minSwaps(string str1, string str2)
{
    // Initialize the count
    int count = 0;
  
    // Iterate the loop with str1 length
    for (int i = 0; i < str1.Length; i++) {
  
        // If any non-equal elements are found
        // increment the counter
        if (str1[i] != str2[i])
            count++;
    }
  
    // If counter is even print the swap
    if (count % 2 == 0)
        Console.WriteLine(count / 2);
    else
        Console.WriteLine("Not Possible");
}
  
// Driver Code
public static void Main()
{
    // Take two input
    string binaryString1 = "1110000";
    string binaryString2 = "0001101";
  
    // Call the method
    minSwaps(binaryString1, binaryString2);
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

输出：

3

时间复杂度： O(n)