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

• Difficulty Level : Hard
• Last Updated : 13 Jun, 2022

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

Time Complexity: Onlogn)

Auxiliary Space: O(n)

My Personal Notes arrow_drop_up
Recommended Articles
Page :