Last remaining character after repeated removal of the first character and flipping of characters of a Binary String

Given a Binary string str of length N, the task is to find the last character removed from the string by repeatedly removing the first character of the string and flipping all the characters of the string if the removed character is ‘0’.

Examples:

Input: str = “1001”
Output: ‘0’
Explanation:
Removing str[0] from the string modifies str to “001”.
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’

Naive Approach: The simplest approach to solve this problem is to iterate over the characters of the string. For every character encountered, remove the first character of the string and check if the removed character was ‘0’ or not. If found to be true, then flip all the characters of the string. Finally, print the character which was removed in the last iteration. Follow the steps below to solve the problem:

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observation:

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].

Follow the steps below to solve the problem:

• If N is equal to 1 then answer is str[0] itself, Otherwise.
• Check if str[N – 2] (For N>=2) is ‘1’ or not. If found to be true, then print str[N – 1].
• Otherwise, print (‘1’ –  str[N – 1] + ‘0’).

Below is the implementation of the above approach:

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`

Javascript

 ``

Output:
`0`

Time Complexity: O(1), The time complexity is O(1), i.e., constant time. This is because the program performs a fixed set of operations on a given input string of a fixed length, regardless of the contents of the string. Therefore, the time complexity is not dependent on the size of the input.
Auxiliary Space: O(1), The space complexity is O(1), i.e., constant space. This is because the program uses a fixed amount of memory to store the input string and some variables for intermediate calculations. The space used by the program does not increase with the size of the input string, and it remains constant for any input string of a fixed length.

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