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 <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;
}

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

<script>
 
      // JavaScript implementation of the approach
       
      // Function that returns true if the
      // current arrangement is valid
      function check(s) {
        for (var i = 0; i + 1 < s.length; ++i)
        {
          if (Math.abs(s[i].charCodeAt(0) -
          s[i + 1].charCodeAt(0)) === 1)
          {
            return false;
          }
        }
        return true;
      }
 
      // Function to rearrange the characters of
      // the string such that no two neighbours
      // in the English alphabets appear together
      function Rearrange(str) {
        // To store the odd and the
        // even positioned characters
        var odd = "",
          even = "";
 
        // Traverse through the array
        for (var i = 0; i < str.length; ++i) {
          if (str[i].charCodeAt(0) % 2 === 0) {
            even += str[i];
          } else {
            odd += str[i];
          }
        }
 
        // Sort both the strings
        odd.split("").sort((a, b) => a - b);
        even.split("").sort((a, b) => a - b);
 
        // Check possibilities
        if (check(odd + even)) {
          document.write(odd + even);
        } else if (check(even + odd)) {
          document.write(even + odd);
        } else {
          document.write(-1);
        }
      }
 
      // Driver code
      var str = "aabcd";
      Rearrange(str);
       
</script>

Output:

bdaac

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