获取给定二进制字符串所需的最少操作数

给定一个长度为N的二进制字符串S ，任务是通过最少的操作次数从一个长度为N的字符串获得S ，比如T ，长度为N ，仅由零组成。每个操作都涉及从字符串S 中选择任何索引i并翻转字符串T 的索引[i, N – 1]处的所有位。
例子：

Input: S = “101” 
Output: 3 
Explanation: 
“000” -> “111” -> “100” -> “101”. 
Therefore, the minimum number of steps required is 3.
Input: S = “10111” 
Output: 3 
Explanation: 
“00000” -> “11111” -> “10000” -> “10111”. 
Therefore, the minimum number of steps required is 3. 
 

方法：
这个想法是在给定的字符串S 中找到第一次出现的1并在该索引处执行给定的操作。在此步骤之后，对于特定索引处S和T字符的每个不匹配，重复该操作。
请按照以下步骤操作：

1. 迭代S并标记1的第一次出现。
2. 初始化两个变量，比如last和ans ，其中last存储执行最后一个操作的字符（ 0 或 1 ），并且ans保持所需的最小步骤数的计数。
3. 从第一次出现的1迭代到字符串。
4. 如果当前字符为0且last = 1 ，则增加ans的计数，并设置last = 0 。
5. 否则，如果last = 0且当前字符为1 ，则设置last = 1 。
6. 返回ans的最终值。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the minimum
// number of operations required
// to obtain the string s
int minOperations(string s)
{
    int n = s.size();
    int pos = -1;
 
    // Iterate the string s
    for (int i = 0; i < s.size(); i++) {
 
        // If first occurrence of 1
        // is found
        if (s[i] == '1') {
 
            // Mark the index
            pos = i;
            break;
        }
    }
 
    // Base case: If no 1 occurred
    if (pos == -1) {
 
        // No operations required
        return 0;
    }
 
    // Stores the character
    // for which last operation
    // was performed
    int last = 1;
 
    // Stores minimum number
    // of operations
    int ans = 1;
 
    // Iterate from pos to n
    for (int i = pos + 1; i < n; i++) {
 
        // Check if s[i] is 0
        if (s[i] == '0') {
 
            // Check if last operation was
            // performed because of 1
            if (last == 1) {
                ans++;
 
                // Set last to 0
                last = 0;
            }
        }
        else {
 
            if (last == 0) {
 
                // Check if last operation was
                // performed because of 0
                ans++;
 
                // Set last to 1
                last = 1;
            }
        }
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
int main()
{
    string s = "10111";
 
    cout << minOperations(s);
    return 0;
}

Java

// Java program to implement the
// above approach
import java.util.*;
 
class GFG{
 
// Function to find the minimum
// number of operations required
// to obtain the string s
static int minOperations(String s)
{
    int n = s.length();
    int pos = -1;
 
    // Iterate the string s
    for(int i = 0; i < s.length(); i++)
    {
         
        // If first occurrence of 1
        // is found
        if (s.charAt(i) == '1')
        {
             
            // Mark the index
            pos = i;
            break;
        }
    }
 
    // Base case: If no 1 occurred
    if (pos == -1)
    {
 
        // No operations required
        return 0;
    }
 
    // Stores the character
    // for which last operation
    // was performed
    int last = 1;
 
    // Stores minimum number
    // of operations
    int ans = 1;
 
    // Iterate from pos to n
    for(int i = pos + 1; i < n; i++)
    {
         
        // Check if s[i] is 0
        if (s.charAt(i) == '0')
        {
 
            // Check if last operation was
            // performed because of 1
            if (last == 1)
            {
                ans++;
 
                // Set last to 0
                last = 0;
            }
        }
        else
        {
            if (last == 0)
            {
 
                // Check if last operation was
                // performed because of 0
                ans++;
 
                // Set last to 1
                last = 1;
            }
        }
    }
     
    // Return the answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "10111";
     
    System.out.println(minOperations(s));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
 
# Function to find the minimum
# number of operations required
# to obtain the string s
def minOperations(s):
 
    n = len(s)
    pos = -1
 
    # Iterate the string s
    for i in range(len(s)):
 
        # If first occurrence of 1
        # is found
        if (s[i] == '1'):
 
            # Mark the index
            pos = i
            break
 
    # Base case: If no 1 occurred
    if (pos == -1):
 
        # No operations required
        return 0
 
    # Stores the character
    # for which last operation
    # was performed
    last = 1
 
    # Stores minimum number
    # of operations
    ans = 1
 
    # Iterate from pos to n
    for i in range(pos + 1, n):
 
        # Check if s[i] is 0
        if (s[i] == '0'):
 
            # Check if last operation was
            # performed because of 1
            if (last == 1):
                ans += 1
 
                # Set last to 0
                last = 0
         
        else:
 
            if (last == 0):
 
                # Check if last operation was
                # performed because of 0
                ans += 1
 
                # Set last to 1
                last = 1
 
    # Return the answer
    return ans
 
# Driver Code
if __name__ == "__main__":
     
    s = "10111"
 
    print(minOperations(s))
 
# This code is contributed by chitranayal

C#

// C# program to implement the
// above approach
using System;
class GFG{
  
// Function to find the minimum
// number of operations required
// to obtain the string s
static int minOperations(String s)
{
    int n = s.Length;
    int pos = -1;
  
    // Iterate the string s
    for(int i = 0; i < s.Length; i++)
    {
          
        // If first occurrence of 1
        // is found
        if (s[i] == '1')
        {
              
            // Mark the index
            pos = i;
            break;
        }
    }
  
    // Base case: If no 1 occurred
    if (pos == -1)
    {
  
        // No operations required
        return 0;
    }
  
    // Stores the character
    // for which last operation
    // was performed
    int last = 1;
  
    // Stores minimum number
    // of operations
    int ans = 1;
  
    // Iterate from pos to n
    for(int i = pos + 1; i < n; i++)
    {
          
        // Check if s[i] is 0
        if (s[i] == '0')
        {
  
            // Check if last operation was
            // performed because of 1
            if (last == 1)
            {
                ans++;
  
                // Set last to 0
                last = 0;
            }
        }
        else
        {
            if (last == 0)
            {
  
                // Check if last operation was
                // performed because of 0
                ans++;
  
                // Set last to 1
                last = 1;
            }
        }
    }
      
    // Return the answer
    return ans;
}
  
// Driver code
public static void Main(string[] args)
{
    String s = "10111";
      
    Console.Write(minOperations(s));
}
}
  
// This code is contributed by rock_cool

Javascript

    // Javascript Program to implement
    // the above approach
     
    // Function to find the minimum
    // number of operations required
    // to obtain the string s
    function minOperations(s)
    {
        let n = s.length;
        let pos = -1;
 
        // Iterate the string s
        for (let i = 0; i < s.length; i++) {
 
            // If first occurrence of 1
            // is found
            if (s[i] == '1') {
 
                // Mark the index
                pos = i;
                break;
            }
        }
 
        // Base case: If no 1 occurred
        if (pos == -1) {
 
            // No operations required
            return 0;
        }
 
        // Stores the character
        // for which last operation
        // was performed
        let last = 1;
 
        // Stores minimum number
        // of operations
        let ans = 1;
 
        // Iterate from pos to n
        for (let i = pos + 1; i < n; i++) {
 
            // Check if s[i] is 0
            if (s[i] == '0') {
 
                // Check if last operation was
                // performed because of 1
                if (last == 1) {
                    ans++;
 
                    // Set last to 0
                    last = 0;
                }
            }
            else {
 
                if (last == 0) {
 
                    // Check if last operation was
                    // performed because of 0
                    ans++;
 
                    // Set last to 1
                    last = 1;
                }
            }
        }
 
        // Return the answer
        return ans;
    }
     
    let s = "10111";
    document.write(minOperations(s));
 
// This code is contributed by suresh07.

输出：

3

时间复杂度： O(N)
辅助空间： O(1)

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