# 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:**

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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(N^{2})**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 <bits/stdc++.h>` `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

`<script>` ` ` `// JavaScript program to implement` ` ` `// the above approach` ` ` `// Function to find the last removed` ` ` `// character from the string` ` ` `function` `lastRemovedCharacter(str) {` ` ` `// Stores length of the string` ` ` `var` `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` ` ` `var` `str = ` `"10010"` `;` ` ` `document.write(lastRemovedCharacter(str));` ` ` `</script>` |

**Output:**

0

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