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# Rearrange the characters of the string such that no two adjacent characters are consecutive English alphabets

• Difficulty Level : Hard
• Last Updated : 18 May, 2021

Given string str of size N consists of lower-case English alphabets. The task is to find the arrangement of the characters of the string such that no two adjacent characters are neighbors in English alphabets. In case of multiple answers print any of them. If no such arrangement is possible then print -1.
Examples:

Input: str = “aabcd”
Output: bdaac
No two adjacent characters are neighbours in English alphabets.
Input: str = “aab”
Output: -1

Approach: Traverse through the string and store all odd positioned characters in a string odd and even positioned characters in another string even i.e. every two consecutive characters in both the strings will have an absolute difference in ASCII values of at least 2. Then sort both the strings. Now, if any of the concatenation (even + odd) or (odd + even) is valid then print the valid arrangement else it is not possible to rearrange the string in the required way.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function that returns true if the``// current arrangement is valid``bool` `check(string s)``{``    ``for` `(``int` `i = 0; i + 1 < s.size(); ++i)``        ``if` `(``abs``(s[i] - s[i + 1]) == 1)``            ``return` `false``;``    ``return` `true``;``}` `// Function to rearrange the characters of``// the string such that no two neighbours``// in the English alphabets appear together``void` `Rearrange(string str)``{``    ``// To store the odd and the``    ``// even positioned characters``    ``string odd = ``""``, even = ``""``;` `    ``// Traverse through the array``    ``for` `(``int` `i = 0; i < str.size(); ++i) {``        ``if` `(str[i] % 2 == 0)``            ``even += str[i];``        ``else``            ``odd += str[i];``    ``}` `    ``// Sort both the strings``    ``sort(odd.begin(), odd.end());``    ``sort(even.begin(), even.end());` `    ``// Check possibilities``    ``if` `(check(odd + even))``        ``cout << odd + even;``    ``else` `if` `(check(even + odd))``        ``cout << even + odd;``    ``else``        ``cout << -1;``}` `// Driver code``int` `main()``{``    ``string str = ``"aabcd"``;` `    ``Rearrange(str);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `    ``// Function that returns true if the``    ``// current arrangement is valid``    ``static` `boolean` `check(String s)``    ``{``        ``for` `(``int` `i = ``0``; i + ``1` `< s.length(); ++i)``        ``{``            ``if` `(Math.abs(s.charAt(i) -``                         ``s.charAt(i + ``1``)) == ``1``)``            ``{``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``// Function to rearrange the characters of``    ``// the string such that no two neighbours``    ``// in the English alphabets appear together``    ``static` `void` `Rearrange(String str)``    ``{``        ` `        ``// To store the odd and the``        ``// even positioned characters``        ``String odd = ``""``, even = ``""``;` `        ``// Traverse through the array``        ``for` `(``int` `i = ``0``; i < str.length(); ++i)``        ``{``            ``if` `(str.charAt(i) % ``2` `== ``0``)``            ``{``                ``even += str.charAt(i);``            ``}``            ``else``            ``{``                ``odd += str.charAt(i);``            ``}``        ``}` `        ``// Sort both the strings``        ``odd = sort(odd);``        ``even = sort(even);` `        ``// Check possibilities``        ``if` `(check(odd + even))``        ``{``            ``System.out.print(odd + even);``        ``}``        ``else` `if` `(check(even + odd))``        ``{``            ``System.out.print(even + odd);``        ``}``        ``else``        ``{``            ``System.out.print(-``1``);``        ``}``    ``}``    ` `    ``// Method to sort a string alphabetically``    ``public` `static` `String sort(String inputString)``    ``{``        ``// convert input string to char array``        ``char` `tempArray[] = inputString.toCharArray();` `        ``// sort tempArray``        ``Arrays.sort(tempArray);` `        ``// return new sorted string``        ``return` `new` `String(tempArray);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"aabcd"``;` `        ``Rearrange(str);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach` `# Function that returns true if the``# current arrangement is valid``def` `check(s):` `    ``for` `i ``in` `range``(``len``(s) ``-` `1``):``        ``if` `(``abs``(``ord``(s[i]) ``-``                ``ord``(s[i ``+` `1``])) ``=``=` `1``):``            ``return` `False``    ``return` `True` `# Function to rearrange the characters``# of the such that no two neighbours``# in the English alphabets appear together``def` `Rearrange(``Str``):` `    ``# To store the odd and the``    ``# even positioned characters``    ``odd, even ``=` `"``","``"` `    ``# Traverse through the array``    ``for` `i ``in` `range``(``len``(``Str``)):``        ``if` `(``ord``(``Str``[i]) ``%` `2` `=``=` `0``):``            ``even ``+``=` `Str``[i]``        ``else``:``            ``odd ``+``=` `Str``[i]` `    ``# Sort both the Strings``    ``odd ``=` `sorted``(odd)``    ``even ``=` `sorted``(even)` `    ``# Check possibilities``    ``if` `(check(odd ``+` `even)):``        ``print``("".join(odd ``+` `even))``    ``elif` `(check(even ``+` `odd)):``        ``print``("".join(even ``+` `odd))``    ``else``:``        ``print``(``-``1``)` `# Driver code``Str` `=` `"aabcd"` `Rearrange(``Str``)` `# This code is contributed``# by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `    ``// Function that returns true if the``    ``// current arrangement is valid``    ``static` `Boolean check(String s)``    ``{``        ``for` `(``int` `i = 0; i + 1 < s.Length; ++i)``        ``{``            ``if` `(Math.Abs(s[i] -``                         ``s[i + 1]) == 1)``            ``{``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``// Function to rearrange the characters of``    ``// the string such that no two neighbours``    ``// in the English alphabets appear together``    ``static` `void` `Rearrange(String str)``    ``{``        ` `        ``// To store the odd and the``        ``// even positioned characters``        ``String odd = ``""``, even = ``""``;` `        ``// Traverse through the array``        ``for` `(``int` `i = 0; i < str.Length; ++i)``        ``{``            ``if` `(str[i] % 2 == 0)``            ``{``                ``even += str[i];``            ``}``            ``else``            ``{``                ``odd += str[i];``            ``}``        ``}` `        ``// Sort both the strings``        ``odd = sort(odd);``        ``even = sort(even);` `        ``// Check possibilities``        ``if` `(check(odd + even))``        ``{``            ``Console.Write(odd + even);``        ``}``        ``else` `if` `(check(even + odd))``        ``{``            ``Console.Write(even + odd);``        ``}``        ``else``        ``{``            ``Console.Write(-1);``        ``}``    ``}``    ` `    ``// Method to sort a string alphabetically``    ``public` `static` `String sort(String inputString)``    ``{``        ``// convert input string to char array``        ``char` `[]tempArray = inputString.ToCharArray();` `        ``// sort tempArray``        ``Array.Sort(tempArray);` `        ``// return new sorted string``        ``return` `new` `String(tempArray);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String str = ``"aabcd"``;` `        ``Rearrange(str);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`bdaac`

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