去除重复的第一个字符后，最后剩下的字符和二进制字符串的字符翻转

鉴于长度为N的二进制字符串str，任务是通过重复删除字符串的第一个字符和翻转字符串的所有字符，如果删除的字符是“0”，找到最后一个字符从字符串中删除。

例子：

Input: str = “1001” 
Output: ‘0’ 
Explanation: 
Removing str[0] from the string modifies str to “00”. 
Removing str[0] from the string modifies str to “10”. 
Removing str[0] from the string modifies str to “0”. 
Since the last character removed was ‘0’, the required output is ‘0’.

Input: str = “10010” 
Output: ‘0’

原始的方法：要解决这个问题，最简单的方法是遍历字符串的字符。对于遇到的每个字符，请删除字符串的第一个字符，然后检查所删除的字符为“ 0” 。如果发现是真的，则翻转字符串的所有字符。最后，打印在上一次迭代中删除的字符。请按照以下步骤解决问题：

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：可以基于以下观察来优化上述方法：

Case 1: 
Suppose str = “XXXXXXX0X”, where X is either ‘0’ or ‘1’. 
If the character str[N – 2] is flipped, then str[N – 1] must be flipped. 
Therefore, if str[N – 2] is ‘0’, then last removed character will be (‘1’ – str[N – 1] + ‘0’).
Case 2: 
Suppose str = “XXXXXXX1X”, where X is either ‘0’ or ‘1. 
If the character str[N – 2] is flipped, then str[N – 1] must be flipped. 
Therefore, if str[N – 2] is ‘1’, then the last removed character will be str[N – 1]. 
 

请按照以下步骤解决问题：

• 检查str [N – 2]是否为‘1’ 。如果发现为真，则打印str [N – 1]。
• 否则，打印(’1’– str [N – 1] +’0’) 。

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 last removed
// character from the string
char lastRemovedCharacter(string str)
{
    // Stores length of the string
    int n = str.length();
 
    // Base Case:
    if (n == 1)
        return str[0];
 
    // If the second last
    // character is '0'
    if (str[n - 2] == '0') {
 
        return ('1' - str[n - 1] + '0');
    }
 
    // If the second last
    // character is '1'
    else
        return str[n - 1];
}
 
// Driver Code
int main()
{
    string str = "10010";
    cout << lastRemovedCharacter(str);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to find the last removed
// character from the String
static char lastRemovedCharacter(char []str)
{
    // Stores length of the String
    int n = str.length;
 
    // Base Case:
    if (n == 1)
        return str[0];
 
    // If the second last
    // character is '0'
    if (str[n - 2] == '0') {
 
        return (char)('1' - str[n - 1] + '0');
    }
 
    // If the second last
    // character is '1'
    else
        return str[n - 1];
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "10010";
    System.out.print(lastRemovedCharacter(str.toCharArray()));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
  
# Function to find the last removed
# character from the string
def lastRemovedCharacter(str):
     
    # Stores length of the string
    n = len(str)
  
    # Base Case:
    if (n == 1):
        return ord(str[0])
  
    # If the second last
    # character is '0'
    if (str[n - 2] == '0'):
        return (ord('1') -
                ord(str[n - 1]) +
                ord('0'))
  
    # If the second last
    # character is '1'
    else:
        return ord(str[n - 1])
  
# Driver Code
if __name__ == '__main__':
     
    str = "10010"
     
    print(chr(lastRemovedCharacter(str)))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach 
using System;
    
class GFG{
    
// Function to find the last removed
// character from the String
static char lastRemovedCharacter(char []str)
{
     
    // Stores length of the String
    int n = str.Length;
  
    // Base Case:
    if (n == 1)
        return str[0];
  
    // If the second last
    // character is '0'
    if (str[n - 2] == '0')
    {
        return (char)('1' - str[n - 1] + '0');
    }
  
    // If the second last
    // character is '1'
    else
        return str[n - 1];
}
    
// Driver Code
public static void Main()
{
    string str = "10010";
     
    Console.Write(lastRemovedCharacter(
        str.ToCharArray()));
}
}
 
// This code is contributed by code_hunt

输出：

0

时间复杂度： O(1)
辅助空间： O(1)