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 validbool 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 togethervoid 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 codeint main(){ string str = "aabcd"; Rearrange(str); return 0;} |
Java
// Java implementation of the approachimport 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 validdef 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 togetherdef 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 codeStr = "aabcd"Rearrange(Str)# This code is contributed# by Mohit Kumar |
C#
// C# implementation of the approachusing 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
Time Complexity: Onlogn)
Auxiliary Space: O(n)
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