Related Articles

# Check if it is possible to rearrange a binary string with alternate 0s and 1s

• Difficulty Level : Basic
• Last Updated : 02 Jul, 2021

Given a binary string of length, at least two. We need to check if it is possible to rearrange a binary string such that there alternate 0’s and 1’s. If possible, then the output is YES, otherwise the output is NO.

Examples:

Input : 1011
Output : NO
We can’t rearrange the string such that it has alternate 0s and 1s.

Input : 1100
Output : YES
There are exactly two ways to rearrange the string, and they are 0101 or 1010 .

We can place all 0’s in an even position and all 1’s in an odd position or we can place all 0’s in an odd position and all 1’s in an even position. If the length of the string is even, then to satisfy the given condition, the count of 1’s and 0’s must be equal. If the length of the string is odd then to satisfy the given condition, the absolute difference in the count
of 1’s and 0’s must be one.

Below is the implementation of the above approach:

## C++

 `// CPP program to check if we can rearrange a``// string such that it has alternate 0s and 1s.``#include ``using` `namespace` `std;` `// function to check the binary string``bool` `is_possible(string s)``{``    ``// length of string``    ``int` `l = s.length();` `    ``int` `one = 0, zero = 0;` `    ``for` `(``int` `i = 0; i < l; i++) {` `        ``// count zero's``        ``if` `(s[i] == ``'0'``)``            ``zero++;` `        ``// count one's``        ``else``            ``one++;``    ``}` `    ``// if length is even``    ``if` `(l % 2 == 0)``        ``return` `(one == zero);` `    ``// if length is odd``    ``else``        ``return` `(``abs``(one - zero) == 1);``}` `// Driver code``int` `main()``{``    ``string s = ``"100110"``;``    ``if` `(is_possible(s))``      ``cout << ``"Yes"``;``    ``else``      ``cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program to check if we can rearrange a``// string such that it has alternate 0s and 1s.``import` `java.lang.Math;` `public` `class` `GfG{``    ` `    ``// function to check the binary string``    ``public` `static` `boolean` `is_possible(String s)``    ``{``        ``// length of string``        ``int` `l = s.length();``      ` `        ``int` `one = ``0``, zero = ``0``;``      ` `        ``for` `(``int` `i = ``0``; i < l; i++) {``      ` `            ``// count zero's``            ``if` `(s.charAt(i) == ``'0'``)``                ``zero++;``      ` `            ``// count one's``            ``else``                ``one++;``        ``}``      ` `        ``// if length is even``        ``if` `(l % ``2` `== ``0``) ``            ``return` `(one == zero);``      ` `        ``// if length is odd``        ``else``            ``return` `(Math.abs(one - zero) == ``1``);``    ``}` `    ``public` `static` `void` `main(String []args){``        ` `        ``String s = ``"100110"``;``        ``if` `(is_possible(s))``          ``System.out.println(``"Yes"``);``        ``else``          ``System.out.println(``"No"``);``    ``}``}` `// This code is contributed by Rituraj Jain`

## Python 3

 `# Python3 program to check if``# we can rearrange a``# string such that it has alternate``# 0s and 1s.` `# function to check the binary string``def` `is_possible(s):` `    ``# length of string``    ``l ``=` `len``(s)` `    ``one ``=` `0``    ``zero ``=` `0` `    ``for` `i ``in` `range``(``0``,l) :` `        ``# count zero's``        ``if` `(s[i] ``=``=` `'0'``):``            ``zero ``+``=` `1` `        ``# count one's``        ``else``:``            ``one ``+``=` `1` `    ``# if length is even``    ``if` `(l ``%` `2` `=``=` `0``) :``        ``return` `(one ``=``=` `zero)` `    ``# if length is odd``    ``else``:``        ``return` `(``abs``(one ``-` `zero) ``=``=` `1``)` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``s ``=` `"100110"``    ``if` `(is_possible(s)):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)` `# This code is contributed by``# ChitraNayal`

## C#

 `// C# program to check if we can rearrange a``// string such that it has alternate 0s and 1s.``using` `System;` `class` `GfG``{``    ` `    ``// function to check the binary string``    ``public` `static` `bool` `is_possible(String s)``    ``{``        ``// length of string``        ``int` `l = s.Length;``        ` `        ``int` `one = 0, zero = 0;``        ` `        ``for` `(``int` `i = 0; i < l; i++)``        ``{``        ` `            ``// count zero's``            ``if` `(s[i] == ``'0'``)``                ``zero++;``        ` `            ``// count one's``            ``else``                ``one++;``        ``}``        ` `        ``// if length is even``        ``if` `(l % 2 == 0)``            ``return` `(one == zero);``        ` `        ``// if length is odd``        ``else``            ``return` `(Math.Abs(one - zero) == 1);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String []args)``    ``{``        ` `        ``String s = ``"100110"``;``        ``if` `(is_possible(s))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by Rajput-Ji`

## PHP

 ``

## Javascript

 ``
Output:
`Yes`

Time Complexity: O(l), where l is the length of the binary string.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up