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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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 



















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Output:


bdaac






          
          
          
          

            

    

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