将给定的二进制字符串转换为另一个字符串所需的最小子字符串翻转

给定两个二进制字符串A和B ，任务是查找从A的第一个字符开始的子字符串需要翻转的最小次数，即将1 s转换为0 s，将0 s转换为1 s，以将A转换为B．

例子：

Input: A = “0010”, B = “1011”
Output; 3
Explanation:
Step 1: Flip the entire string A. Therefore, A becomes “1101” .
Step 2: Flip the substring {A[0], A[2]}. Therefore, A becomes “0011” .
Step 3: Flip A[0]. Therefore, A becomes “1011” which is equal to B.
Therefore, the minimum number of operations required is 3.

Input: A = “1010101”, B = “0011100”
Output: 5
Explanation:
Step 1: Flip the entiThehrefore, A becomes “0101010″
Step 2: Flip substring {A[0], A[5]}. Therefore, A becomes “1010100″
Step 3: Flip the substring {A[0], A[3]}. Therefore, A becomes “0101100″
Step 4: Flip the substring {A[0], A[2]}. Therefore, A becomes “1011100″
Step 5: Flip A[0]. Therefore, A becomes “0011100” which is equal to B.
Therefore, the minimum number of operations required is 5.

的方法：我们的想法是初始化，保持最后索引在该字符在A是从在B点的字符不同的变量。然后从第一个索引到最后一个索引取反A。重复直到两个字符串相等。请按照以下步骤解决问题：

• 初始化一个变量last_index ，该变量包含A和B中字符不同的最后一个索引。
• 将字符串A从第一个索引取反到last_index，并增加步数。
• 重复上述步骤，直到字符串A等于字符串B为止。
• 执行完两个字符串相同的字符串后，打印步数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function that finds the minimum
// number of operations required such
// that string A and B are the same
void findMinimumOperations(string a,
                           string b)
{
 
    // Stores the count of steps
    int step = 0;
 
    // Stores the last index whose
    // bits are not same
    int last_index;
 
    // Iterate until both string
    // are unequal
    while (a != b) {
 
        // Check till end of string to
        // find righmost unequals bit
        for (int i = 0;
             i < a.length(); i++) {
 
            // Update the last index
            if (a[i] != b[i]) {
                last_index = i;
            }
        }
 
        // Flipping characters up
        // to the last index
        for (int i = 0;
             i <= last_index; i++) {
 
            // Flip the bit
            a[i] = (a[i] == '0') ? '1' : '0';
        }
 
        // Increasing steps by one
        step++;
    }
 
    // Print the count of steps
    cout << step;
}
 
// Driver Code
int main()
{
    // Given strings A and B
    string A = "101010", B = "110011";
 
    // Function Call
    findMinimumOperations(A, B);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function that finds the minimum
// number of operations required such
// that String A and B are the same
static void findMinimumOperations(char[] a,
                                  char[] b)
{
  // Stores the count of steps
  int step = 0;
 
  // Stores the last index whose
  // bits are not same
  int last_index = 0;
 
  // Iterate until both String
  // are unequal
  while (!Arrays.equals(a, b))
  {
    // Check till end of String to
    // find righmost unequals bit
    for (int i = 0;
             i < a.length; i++)
    {
      // Update the last index
      if (a[i] != b[i])
      {
        last_index = i;
      }
    }
 
    // Flipping characters up
    // to the last index
    for (int i = 0;
             i <= last_index; i++)
    {
 
      // Flip the bit
      a[i] = (a[i] == '0') ?
              '1' : '0';
    }
    // Increasing steps by one
    step++;
  }
 
  // Print the count of steps
  System.out.print(step);
}
 
// Driver Code
public static void main(String[] args)
{
    // Given Strings A and B
    String A = "101010",
           B = "110011";
 
    // Function Call
    findMinimumOperations(A.toCharArray(),
                          B.toCharArray());
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function that finds the minimum
# number of operations required such
# that string A and B are the same
def findMinimumOperations(a, b):
     
    # Stores the count of steps
    step = 0
 
    # Stores the last index whose
    # bits are not same
    last_index = 0
 
    # Iterate until both string
    # are unequal
    while (a != b):
        a = [i for i in a]
         
        # Check till end of string to
        # find righmost unequals bit
        for i in range(len(a)):
             
            # Update the last index
            if (a[i] != b[i]):
                last_index = i
 
        # Flipping characters up
        # to the last index
        for i in range(last_index + 1):
             
            # Flip the bit
            if (a[i] == '0'):
                a[i] = '1'
            else:
                a[i] = '0'
                 
        a = "".join(a)       
 
        # Increasing steps by one
        step += 1
 
    # Print the count of steps
    print(step)
 
# Driver Code
if __name__ == '__main__':
     
    # Given strings A and B
    A = "101010"
    B = "110011"
 
    # Function Call
    findMinimumOperations(A, B)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System;
 
class GFG{
 
// Function that finds the minimum
// number of operations required such
// that string A and B are the same
static void findMinimumOperations(string a,
                                  string b)
{
   
  // Stores the count of steps
  int step = 0;
 
  // Stores the last index whose
  // bits are not same
  int last_index = 0;
 
  // Iterate until both string
  // are unequal
  while (a.Equals(b) == false)
  {
     
    // Check till end of string to
    // find righmost unequals bit
    for(int i = 0; i < a.Length; i++)
    {
       
      // Update the last index
      if (a[i] != b[i])
      {
        last_index = i;
      }
    }
 
    // Flipping characters up
    // to the last index
    char[] ch = a.ToCharArray();
    for(int i = 0;
            i <= last_index; i++)
    {
       
      // Flip the bit
      if (ch[i] == '0')
      {
        ch[i] = '1';
      }
      else
      {
        ch[i] = '0';
      }
    }
    a = new string(ch);
     
    // Increasing steps by one
    step++;
  }
 
  // Print the count of steps
  Console.WriteLine(step);
}
 
// Driver Code
public static void Main()
{
   
  // Given strings A and B
  string A = "101010";
  string B = "110011";
 
  // Function Call
  findMinimumOperations(A,B);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

输出：

4

时间复杂度： O(N ² )
辅助空间： O(1)