# Smallest string obtained by removing all occurrences of 01 and 11 from Binary String | Set 2

Given a binary string **S **of length **N**, 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 = “1010”Output:2Explanation:Removal of substring “01” modifies string S to “10”.

Input:S = “111”Output:1

**Stack****-based Approach:** Refer to the previous article to find the length of the smallest string possible by given operations.**Time Complexity:** O(N)**Auxiliary Space:** O(N)

**Spac-Optimized Approach: **The above approach can be space-optimized by only storing the length of the characters not removed. Follow the steps below to solve the problem:

- Initialize a variable, say
**st**as**0,**to store the length of the smallest string possible. - Iterate over the characters of the string
**S**and perform the following steps:- If
**st**is greater than**0**and**S[i]**is equal to ‘**1**‘, then pop the last element by decrementing**st**by**1**. - Otherwise, increment
**st**by**1.**

- If
- Finally, after completing the above steps, print the answer obtained in
**st**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the length of` `// the smallest string possible by` `// removing substrings "01" and "11"` `int` `shortestString(string S, ` `int` `N)` `{` ` ` `// Stores the length of` ` ` `// the smallest string` ` ` `int` `st = 0;` ` ` `// Traverse the string S` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// If st is greater` ` ` `// than 0 and S[i] is '1'` ` ` `if` `(st && S[i] == ` `'1'` `) {` ` ` `// Delete the last` ` ` `// character and` ` ` `// decrement st by 1` ` ` `st--;` ` ` `}` ` ` `// Otherwise` ` ` `else` `{` ` ` `// Increment st by 1` ` ` `st++;` ` ` `}` ` ` `}` ` ` `// Return the answer in st` ` ` `return` `st;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Input` ` ` `string S = ` `"1010"` `;` ` ` `int` `N = S.length();` ` ` `// Function call` ` ` `cout << shortestString(S, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `public` `class` `GFG_JAVA {` ` ` `// Function to find the length of` ` ` `// the smallest string possible by` ` ` `// removing substrings "01" and "11"` ` ` `static` `int` `shortestString(String S, ` `int` `N)` ` ` `{` ` ` `// Stores the length of` ` ` `// the smallest string` ` ` `int` `st = ` `0` `;` ` ` `// Traverse the string S` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` `// If st is greater` ` ` `// than 0 and S[i] is '1'` ` ` `if` `(st > ` `0` `&& S.charAt(i) == ` `'1'` `) {` ` ` `// Delete the last` ` ` `// character and` ` ` `// decrement st by 1` ` ` `st--;` ` ` `}` ` ` `// Otherwise` ` ` `else` `{` ` ` `// Increment st by 1` ` ` `st++;` ` ` `}` ` ` `}` ` ` `// Return the answer in st` ` ` `return` `st;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Input` ` ` `String S = ` `"1010"` `;` ` ` `int` `N = S.length();` ` ` `// Function call` ` ` `System.out.println(shortestString(S, N));` ` ` `}` `}` `// This code is contributed by abhinavjain194` |

## Python3

`# Python3 program for the above approach` `# Function to find the length of` `# the smallest string possible by` `# removing substrings "01" and "11"` `def` `shortestString(S, N) :` ` ` ` ` `# Stores the length of` ` ` `# the smallest string` ` ` `st ` `=` `0` `;` ` ` `# Traverse the string S` ` ` `for` `i ` `in` `range` `(N) :` ` ` `# If st is greater` ` ` `# than 0 and S[i] is '1'` ` ` `if` `(st ` `and` `S[i] ` `=` `=` `'1'` `) :` ` ` `# Delete the last` ` ` `# character and` ` ` `# decrement st by 1` ` ` `st ` `-` `=` `1` `;` ` ` `# Otherwise` ` ` `else` `:` ` ` `# Increment st by 1` ` ` `st ` `+` `=` `1` `;` ` ` `# Return the answer in st` ` ` `return` `st;` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `# Input` ` ` `S ` `=` `"1010"` `;` ` ` `N ` `=` `len` `(S);` ` ` `# Function call` ` ` `print` `(shortestString(S, N));` ` ` ` ` `# This code is contributed by AnkThon` |

## C#

`// C# program for the above approach` `using` `System;` `public` `class` `GFG_JAVA {` ` ` `// Function to find the length of` ` ` `// the smallest string possible by` ` ` `// removing substrings "01" and "11"` ` ` `static` `int` `shortestString(` `string` `S, ` `int` `N)` ` ` `{` ` ` ` ` `// Stores the length of` ` ` `// the smallest string` ` ` `int` `st = 0;` ` ` `// Traverse the string S` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// If st is greater` ` ` `// than 0 and S[i] is '1'` ` ` `if` `(st > 0 && S[i] == ` `'1'` `) {` ` ` `// Delete the last` ` ` `// character and` ` ` `// decrement st by 1` ` ` `st--;` ` ` `}` ` ` `// Otherwise` ` ` `else` `{` ` ` `// Increment st by 1` ` ` `st++;` ` ` `}` ` ` `}` ` ` `// Return the answer in st` ` ` `return` `st;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main(` `string` `[] args)` ` ` `{` ` ` `// Input` ` ` `string` `S = ` `"1010"` `;` ` ` `int` `N = S.Length;` ` ` `// Function call` ` ` `Console.WriteLine(shortestString(S, N));` ` ` `}` `}` `// This code is contributed by AnkThon` |

**Output:**

2

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

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