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Smallest string obtained by removing all occurrences of 01 and 11 from Binary String
• Last Updated : 03 Nov, 2020

Given a binary string S , the task is to find the smallest string possible by removing all occurrences of substrings “01” and “11”. After removal of any substring, concatenate the remaining parts of the string.

Examples:

Input: S = “0010110”
Output:
Length = 1 String = 0
Explanation: String can be transformed by the following steps:
0010110 → 00110 → 010 → 0.
Since no occurrence of substrings 01 and 11 are remaining, the string “0” is of minimum possible length 1.

Input: S = “0011101111”
Output: Length = 0
Explanation:
String can be transformed by the following steps:
0011101111 → 01101111 → 011011 → 1011 → 11 → “”.

Approach: To solve the problem, the idea is to observe the following cases:

• 01 and 11 mean that ?1 can be removed where ‘?’ can be 1 or 0.
• The final string will always be in the form 1000… or 000…

This problem can be solved by maintaining a Stack while processing the given string S from left to right. If the current binary digit is 0, add it to the stack, if the current binary digit is 1, remove the top bit from the stack. If the stack is empty, then push the current bit to the stack. Follow the below steps to solve the problem:

1. Initialize a Stack to store the minimum possible string.
2. Traverse the given string over the range [0, N – 1].
3. If the stack is empty, push the current binary digit S[i] in the stack.
4. If the stack is not empty and the current bit S[i] is 1 then remove the top bit from the stack.
5. If the current element S[i] is 0 then, push it to the stack.
6. Finally, append all the elements present in the stack from top to bottom and print it as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find minimum` `// length of the given string` `void` `findMinLength(string s, ``int` `n)` `{` `    ``// Initialize a stack` `    ``stack<``int``> st;`   `    ``// Traverse the string` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// If the stack is empty` `        ``if` `(st.empty())`   `            ``// Push the character` `            ``st.push(s[i]);`   `        ``// If the character is 1` `        ``else` `if` `(s[i] == ``'1'``)`   `            ``// Pop the top element` `            ``st.pop();`   `        ``// Otherwise` `        ``else` `            ``// Push the character` `            ``// to the stack` `            ``st.push(s[i]);` `    ``}`   `    ``// Initialize length` `    ``int` `ans = 0;`   `    ``// Append the characters` `    ``// from top to bottom` `    ``vector<``char``> finalStr;`   `    ``// Until Stack is empty` `    ``while` `(!st.empty()) {` `        ``ans++;` `        ``finalStr.push_back(st.top());` `        ``st.pop();` `    ``}`   `    ``// Print the final string size` `    ``cout << ``"Length = "` `<< ans;`   `    ``// If length of the string is not 0` `    ``if` `(ans != 0) {`   `        ``// Print the string` `        ``cout << ``"\nString = "``;` `        ``for` `(``int` `i = 0; i < ans; i++)` `            ``cout << finalStr[i];` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given string` `    ``string S = ``"101010"``;`   `    ``// String length` `    ``int` `N = S.size();`   `    ``// Function call` `    ``findMinLength(S, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to find minimum` `// length of the given string` `static` `void` `findMinLength(String s, ``int` `n)` `{` `    `  `    ``// Initialize a stack` `    ``Stack st = ``new` `Stack<>();  `   `    ``// Traverse the string` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{` `        `  `        ``// If the stack is empty` `        ``if` `(st.empty())`   `            ``// Push the character` `            ``st.push(s.charAt(i));`   `        ``// If the character is 1` `        ``else` `if` `(s.charAt(i) == ``'1'``)`   `            ``// Pop the top element` `            ``st.pop();`   `        ``// Otherwise` `        ``else` `        `  `            ``// Push the character` `            ``// to the stack` `            ``st.push(s.charAt(i));` `    ``}`   `    ``// Initialize length` `    ``int` `ans = ``0``;`   `    ``// Append the characters` `    ``// from top to bottom` `    ``Vector finalStr = ``new` `Vector();`   `    ``// Until Stack is empty` `    ``while` `(st.size() > ``0``)` `    ``{` `        ``ans++;` `        ``finalStr.add(st.peek());` `        ``st.pop();` `    ``}`   `    ``// Print the final string size` `    ``System.out.println(``"Length = "` `+ ans);`   `    ``// If length of the string is not 0` `    ``if` `(ans != ``0``) ` `    ``{` `        `  `        ``// Print the string` `        ``System.out.print(``"String = "``);` `        `  `        ``for``(``int` `i = ``0``; i < ans; i++)` `            ``System.out.print(finalStr.get(i));` `    ``}` `}`   `// Driver Code` `public` `static` `void` `main(String args[])` `{` `    `  `    ``// Given string` `    ``String S = ``"101010"``;`   `    ``// String length` `    ``int` `N = S.length();`   `    ``// Function call` `    ``findMinLength(S, N);` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR`

## Python3

 `# Python3 program for the above approach`   `from` `collections ``import` `deque`   `# Function to find minimum length` `# of the given string` `def` `findMinLength(s, n):` `  `  `    ``# Initialize a stack` `    ``st ``=` `deque()` `    `  `    ``# Traverse the string from` `    ``# left to right` `    ``for` `i ``in` `range``(n):` `      `  `        ``# If the stack is empty,` `        ``# push the character` `        ``if` `(``len``(st) ``=``=` `0``):` `            ``st.append(s[i])` `        `  `        ``# If the character` `        ``# is B, pop from stack` `        ``elif` `(s[i] ``=``=` `'1'``):` `            ``st.pop()` `        `  `        ``# Otherwise, push the` `        ``# character to the stack` `        ``else``:` `            ``st.append(s[i])` `    `  `    ``# Stores resultant string` `    ``ans ``=` `0` `    ``finalStr ``=` `[]` `    ``while` `(``len``(st) > ``0``):` `        ``ans ``+``=` `1` `        ``finalStr.append(st[``-``1``]);` `        ``st.pop()` `        `  `    ``# Print the final string size` `    ``print``(``"The final string size is: "``, ans)` `    `  `    ``# If length is not 0` `    ``if` `(ans ``=``=` `0``):` `        ``print``(``"The final string is: EMPTY"``)` `    `  `    ``# Print the string` `    ``else``:` `        ``print``(``"The final string is: "``, ``*``finalStr)`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``# Given string` `    ``s ``=` `"0010110"` `    `  `    ``# String length` `    ``n ``=` `7` `    `  `    ``# Function Call` `    ``findMinLength(s, n)`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to find minimum` `// length of the given string` `static` `void` `findMinLength(String s, ``int` `n)` `{` `    `  `    ``// Initialize a stack` `    ``Stack<``char``> st = ``new` `Stack<``char``>();  `   `    ``// Traverse the string` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        `  `        ``// If the stack is empty` `        ``if` `(st.Count == 0)` `        `  `            ``// Push the character` `            ``st.Push(s[i]);`   `        ``// If the character is 1` `        ``else` `if` `(s[i] == ``'1'``)`   `            ``// Pop the top element` `            ``st.Pop();`   `        ``// Otherwise` `        ``else` `        `  `            ``// Push the character` `            ``// to the stack` `            ``st.Push(s[i]);` `    ``}`   `    ``// Initialize length` `    ``int` `ans = 0;`   `    ``// Append the characters` `    ``// from top to bottom` `    ``List<``char``> finalStr = ``new` `List<``char``>();`   `    ``// Until Stack is empty` `    ``while` `(st.Count > 0)` `    ``{` `        ``ans++;` `        ``finalStr.Add(st.Peek());` `        ``st.Pop();` `    ``}`   `    ``// Print the readonly string size` `    ``Console.WriteLine(``"Length = "` `+ ans);`   `    ``// If length of the string is not 0` `    ``if` `(ans != 0) ` `    ``{` `        `  `        ``// Print the string` `        ``Console.Write(``"String = "``);` `        `  `        ``for``(``int` `i = 0; i < ans; i++)` `            ``Console.Write(finalStr[i]);` `    ``}` `}`   `// Driver Code` `public` `static` `void` `Main(String []args)` `{` `    `  `    ``// Given string` `    ``String S = ``"101010"``;`   `    ``// String length` `    ``int` `N = S.Length;`   `    ``// Function call` `    ``findMinLength(S, N);` `}` `}`   `// This code is contributed by Amit Katiyar`

Output:

```Length = 2
String = 01

```

Time Complexity: O(N)
Auxiliary Space: O(N)

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