从二进制字符串删除所有字符串(“ 10”或“ 01”)后打印字符串

给定一个仅由0和1组成的二进制字符串str ，任务是在从字符串逐一删除出现的“ 10”和“ 01”后打印字符串。如果字符串空，则输出-1。

例子：

Input: str = “101100”
Output: -1
Explanation:
In the first step, “10” at index 0 and 1 is removed from the string.
101100 -> 1100
In the second step, “10” at index 1 and 2 is removed from the string.
1100 -> 10
Finally, “10” is removed and the string becomes empty.
10 -> NULL

Input: str = “010110100”
Output: 0
Explanation:
In the first step, “01” at index 0 and 1 is removed from the string.
010110100 -> 0110100
In the second step, “01” at index 0 and 1 is removed from the string.
0110100 -> 10100
In the third step, “10” at index 0 and 1 is removed from the string.
10100 -> 100
Finally, “10” is removed and the string becomes “0”.
100 -> 0

观察：在仔细观察，因为给定的字符串是一个二进制字符串，所有字符串可以除多余的0和1名的是目前无法得到与其配对的恭维字符串中被清除。例如：

Let str = “010110100”.
For this string, the number of 0’s are 5 and the number of 1’s are 4.
Now, let’s start removing alternate substrings one by one:

1. 010110100 -> 0110100
2. 0110100 -> 10100
3. 10100 -> 100
4. 100 -> 0

At this point, the string cannot be further reduced.

因此，从上面的示例可以看出，只要字符串中有1和0，就可以减少字符串。

我们已经在上一篇文章中讨论了查找0和1的缺失计数的方法。在这里，我们对先前的方法做了一些修改，以在所有可能的删除之后生成剩余的字符串。

方法：根据以上观察结果，可以得出结论，最终的字符串仅包含多余的1或0 ，这些数字不能与字符串中的任何数字配对。因此，解决此问题的想法是对字符串中0和1的数目进行计数，然后找出两个计数之间的差。此计数表示以较高的值为准，剩余的1或0 。

可以按照以下步骤计算答案：

1. 获取字符串存在的0计数并将其存储在变量中。
2. 获取字符串存在的1的计数并将其存储在另一个变量中。
3. 如果1的计数等于0的计数，则可以减少整个字符串。因此，返回-1 。
4. 如果1的个数大于0的个数，那么最后剩下的那许多1将被保留，无法进一步减少。因此，请将这许多1附加到一个空字符串，然后返回字符串。
5. 同样，如果0的数量大于1的数量，则求出差值，然后将这多个0附加到空字符串并返回。

下面是上述方法的实现：

C++

// C++ program to print the final string
// after removing all the occurrences of
// "10" and "01" from the given binary string
#include 
using namespace std;
  
// Function to print the final string
// after removing all the occurrences of
// "10" and "01" from the given binary string
void finalString(string str)
{
  
    // Variables to store the
    // count of 1's and 0's
    int x = 0, y = 0;
  
    // Variable left will store
    // whether 0's or 1's is left
    // in the final string
    int left;
  
    // Length of the string
    int n = str.length();
  
    // For loop to count the occurrences
    // of 1's and 0's in the string
    for (int i = 0; i < n; i++) {
        if (str[i] == '1')
            x++;
        else
            y++;
    }
  
    // To check if the count of 1's is
    // greater than the count of 0's or not.
    // If x is greater, then those many 1's
    // are printed.
    if (x > y)
        left = 1;
    else
        left = 0;
  
    // Length of the final remaining string
    // after removing all the occurrences
    int length = n - 2 * min(x, y);
  
    // Printing the final string
    for (int i = 0; i < length; i++) {
        cout << left;
    }
}
  
// Driver Code
int main()
{
    string str = "010110100100000";
    finalString(str);
  
    return 0;
}

Java

// Java program to print the final String
// after removing all the occurrences of
// "10" and "01" from the given binary String
import java.util.*;
  
class GFG{
   
// Function to print the final String
// after removing all the occurrences of
// "10" and "01" from the given binary String
static void finalString(String str)
{
   
    // Variables to store the
    // count of 1's and 0's
    int x = 0, y = 0;
   
    // Variable left will store
    // whether 0's or 1's is left
    // in the final String
    int left;
   
    // Length of the String
    int n = str.length();
   
    // For loop to count the occurrences
    // of 1's and 0's in the String
    for (int i = 0; i < n; i++) {
        if (str.charAt(i) == '1')
            x++;
        else
            y++;
    }
   
    // To check if the count of 1's is
    // greater than the count of 0's or not.
    // If x is greater, then those many 1's
    // are printed.
    if (x > y)
        left = 1;
    else
        left = 0;
   
    // Length of the final remaining String
    // after removing all the occurrences
    int length = n - 2 * Math.min(x, y);
   
    // Printing the final String
    for (int i = 0; i < length; i++) {
        System.out.print(left);
    }
}
   
// Driver Code
public static void main(String[] args)
{
    String str = "010110100100000";
    finalString(str);
}
}
  
// This code is contributed by sapnasingh4991

Python3

# Python 3 program to print the final string
# after removing all the occurrences of
# "10" and "01" from the given binary string
  
# Function to print the final string
# after removing all the occurrences of
# "10" and "01" from the given binary string
def finalString(st):
  
    # Variables to store the
    # count of 1's and 0's
    x , y = 0 , 0
  
    # Length of the string
    n = len(st)
  
    # For loop to count the occurrences
    # of 1's and 0's in the string
    for i in range( n):
        if (st[i] == '1'):
            x += 1
        else:
            y += 1
  
    # To check if the count of 1's is
    # greater than the count of 0's or not.
    # If x is greater, then those many 1's
    # are printed.
    if (x > y):
        left = 1
    else:
        left = 0
  
    # Length of the final remaining string
    # after removing all the occurrences
    length = n - 2 * min(x, y);
  
    # Printing the final string
    for i in range(length):
        print(left, end="")
  
  
# Driver Code
if __name__ == "__main__":
    st = "010110100100000"
    finalString(st)
  
# This code is contributed by chitranayal

C#

// C# program to print the readonly String
// after removing all the occurrences of
// "10" and "01" from the given binary String
using System;
  
class GFG{
    
// Function to print the readonly String
// after removing all the occurrences of
// "10" and "01" from the given binary String
static void finalString(String str)
{
    
    // Variables to store the
    // count of 1's and 0's
    int x = 0, y = 0;
    
    // Variable left will store
    // whether 0's or 1's is left
    // in the readonly String
    int left;
    
    // Length of the String
    int n = str.Length;
    
    // For loop to count the occurrences
    // of 1's and 0's in the String
    for (int i = 0; i < n; i++) {
        if (str[i] == '1')
            x++;
        else
            y++;
    }
    
    // To check if the count of 1's is
    // greater than the count of 0's or not.
    // If x is greater, then those many 1's
    // are printed.
    if (x > y)
        left = 1;
    else
        left = 0;
    
    // Length of the readonly remaining String
    // after removing all the occurrences
    int length = n - 2 * Math.Min(x, y);
    
    // Printing the readonly String
    for (int i = 0; i < length; i++) {
        Console.Write(left);
    }
}
    
// Driver Code
public static void Main(String[] args)
{
    String str = "010110100100000";
    finalString(str);
}
}
   
// This code is contributed by 29AjayKumar

输出：

00000

时间复杂度分析：

• for循环对1和0的出现次数进行计数需要O(N)时间，其中N是字符串的长度。
• if语句需要固定的时间。因此，if语句的时间复杂度为O(1) 。
• 在最坏的情况下，当整个字符串只有0或1时，打印最终字符串的循环将花费O(N) 。
• 因此，总时间复杂度为O(N) 。