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Check if any permutation of string is palindrome

Here given code implementation process.

import java.util.HashMap;
/*
    Java Program
    Check if any permutation of string is palindrome
*/
public class Palindrome
{
	public void isPalindromicPermutation(String text)
	{
		// Use to count character frequency
		HashMap < Character, Integer > frequency = 
          new HashMap < Character, Integer > ();
		boolean status = true;
		boolean odd = false;
		// Count frequency of given text
		for (int i = 0; i < text.length(); ++i)
		{
			if (frequency.containsKey(text.charAt(i)))
			{
              	// Increase character frequency
				frequency.put(text.charAt(i), frequency.get(text.charAt(i)) + 1);
			}
			else
			{
              	// Add new character with one frequency
				frequency.put(text.charAt(i), 1);
			}
		}
		// Check that if given string can be palindrome
		for (char info: frequency.keySet())
		{
			if (frequency.get(info) % 2 != 0)
			{
				if (odd == false)
				{
					// Get single character
					odd = true;
				}
				else
				{
					// When have more than two odd character
					status = false;
					break;
				}
			}
		}
		System.out.print("\n Given Text : " + text);
		if (status == true)
		{
			// When permutation is form of palindrome
			System.out.print("\n Yes ");
			// Auxiliary variables
			String left = "";
			String right = "";
			String middle = "";
			int count = 0;
			// Collect of palindromic sequence
			for (char info: frequency.keySet())
			{
				count = frequency.get(info);
              
				while (count > 0)
				{
					if (count == 1)
					{
						middle += info;
						count--;
					}
					else
					{
						left = left + info;
						right = info + right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			System.out.print(" : [" + left + middle + right + "]");
		}
		else
		{
			System.out.print("\n No ");
		}
	}
	public static void main(String[] args)
	{
		Palindrome task = new Palindrome();
		// Test Cases
		task.isPalindromicPermutation("91002233791");
		task.isPalindromicPermutation("abccbbaa");
		task.isPalindromicPermutation("xyziiyyyx");
	}
}

input

 Given Text : 91002233791
 Yes  : [01239793210]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [xyyiziyyx]
// Include header file
#include <iostream>
#include <string>
#include <unordered_map>
using namespace std;

/*
    C++ Program
    Check if any permutation of string is palindrome
*/

class Palindrome
{
	public: void isPalindromicPermutation(string text)
	{
		// Use to count character frequency
		unordered_map < char, int > frequency;
		bool status = true;
		bool odd = false;
		// Count frequency of given text
		for (int i = 0; i < text.length(); ++i)
		{
			if (frequency.find(text[i]) != frequency.end())
			{
				// Increase character frequency
				frequency[text[i]] = frequency[text[i]] + 1;
			}
			else
			{
				// Add new character with one frequency
				frequency[text[i]] = 1;
			}
		}
		// Check that if given string can be palindrome
		for (auto &info: frequency)
		{
			if (info.second % 2 != 0)
			{
				if (odd == false)
				{
					// Get single character
					odd = true;
				}
				else
				{
					// When have more than two odd character
					status = false;
					break;
				}
			}
		}
		cout << "\n Given Text : " << text;
		if (status == true)
		{
			// When permutation is form of palindrome
			cout << "\n Yes ";
			// Auxiliary variables
			string left = "";
			string right = "";
			string middle = "";
			int count = 0;
			// Collect of palindromic sequence
			for (auto &info: frequency)
			{
				count = info.second;
				while (count > 0)
				{
                  
					if (count == 1)
					{
						middle += info.first;
						count--;
					}
					else
					{
						left = left  +  info.first;
						right = info.first  +  right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			cout << " : [" << left + middle + right << "]";
		}
		else
		{
			cout << "\n No ";
		}
	}
};
int main()
{
	Palindrome *task = new Palindrome();
	// Test Cases
	task->isPalindromicPermutation("91002233791");
	task->isPalindromicPermutation("abccbbaa");
	task->isPalindromicPermutation("xyziiyyyx");
	return 0;
}

input

 Given Text : 91002233791
 Yes  : [39102720193]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [yyixzxiyy]
// Include namespace system
using System;
using System.Collections.Generic;
/*
    Csharp Program
    Check if any permutation of string is palindrome
*/
public class Palindrome
{
	public void isPalindromicPermutation(String text)
	{
		// Use to count character frequency
		Dictionary < char, int > frequency = new Dictionary < char, int > ();
		Boolean status = true;
		Boolean odd = false;
		// Count frequency of given text
		for (int i = 0; i < text.Length; ++i)
		{
			if (frequency.ContainsKey(text[i]))
			{
				// Increase character frequency
				frequency[text[i]] = frequency[text[i]] + 1;
			}
			else
			{
				// Add new character with one frequency
				frequency.Add(text[i], 1);
			}
		}
		// Check that if given string can be palindrome
		foreach(KeyValuePair < char, int > info in frequency)
		{
			if (info.Value % 2 != 0)
			{
				if (odd == false)
				{
					// Get single character
					odd = true;
				}
				else
				{
					// When have more than two odd character
					status = false;
					break;
				}
			}
		}
		Console.Write("\n Given Text : " + text);
		if (status == true)
		{
			// When permutation is form of palindrome
			Console.Write("\n Yes ");
			// Auxiliary variables
			String left = "";
			String right = "";
			String middle = "";
			int count = 0;
			// Collect of palindromic sequence
			foreach(KeyValuePair < char, int > info in frequency)
			{
				count = info.Value;
				while (count > 0)
				{
					if (count == 1)
					{
						middle += info.Key;
						count--;
					}
					else
					{
						left = left + info.Key;
						right = info.Key + right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			Console.Write(" : [" + left + middle + right + "]");
		}
		else
		{
			Console.Write("\n No ");
		}
	}
	public static void Main(String[] args)
	{
		Palindrome task = new Palindrome();
		// Test Cases
		task.isPalindromicPermutation("91002233791");
		task.isPalindromicPermutation("abccbbaa");
		task.isPalindromicPermutation("xyziiyyyx");
	}
}

input

 Given Text : 91002233791
 Yes  : [91023732019]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [xyyiziyyx]
<?php
/*
    Php Program
    Check if any permutation of string is palindrome
*/
class Palindrome
{
	public	function isPalindromicPermutation($text)
	{
		// Use to count character frequency
		$frequency = array();
		$status = true;
		$odd = false;
		// Count frequency of given text
		for ($i = 0; $i < strlen($text); ++$i)
		{
			if (array_key_exists($text[$i], $frequency))
			{
				// Increase character frequency
				$frequency[$text[$i]] = $frequency[$text[$i]] + 1;
			}
			else
			{
				// Add new character with one frequency
				$frequency[$text[$i]] = 1;
			}
		}
		// Check that if given string can be palindrome
		foreach($frequency as $key => $value)
		{
			if ($value % 2 != 0)
			{
				if ($odd == false)
				{
					// Get single character
					$odd = true;
				}
				else
				{
					// When have more than two odd character
					$status = false;
					break;
				}
			}
		}
		echo("\n Given Text : ".$text);
		if ($status == true)
		{
			// When permutation is form of palindrome
			echo("\n Yes ");
			// Auxiliary variables
			$left = "";
			$right = "";
			$middle = "";
			$count = 0;
			// Collect of palindromic sequence
			foreach($frequency as $key => $value)
			{
				$count = $value;
				while ($count > 0)
				{
					if ($count == 1)
					{
						$middle = $middle.$key;
						$count--;
					}
					else
					{
						$left = $left.$key;
						$right = $key.$right;
						$count = $count - 2;
					}
				}
			}
			// Display a palindrome sequence
			echo(" : [".$left.$middle.$right.
				"]");
		}
		else
		{
			echo("\n No ");
		}
	}
}

function main()
{
	$task = new Palindrome();
	// Test Cases
	$task->isPalindromicPermutation("91002233791");
	$task->isPalindromicPermutation("abccbbaa");
	$task->isPalindromicPermutation("xyziiyyyx");
}
main();

input

 Given Text : 91002233791
 Yes  : [91023732019]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [xyyiziyyx]
/*
    Node JS Program
    Check if any permutation of string is palindrome
*/
class Palindrome
{
	isPalindromicPermutation(text)
	{
		// Use to count character frequency
		var frequency = new Map();
		var status = true;
		var odd = false;
		// Count frequency of given text
		for (var i = 0; i < text.length; ++i)
		{
			if (frequency.has(text.charAt(i)))
			{
				// Increase character frequency
				frequency.set(text.charAt(i), frequency.get(text.charAt(i)) + 1);
			}
			else
			{
				// Add new character with one frequency
				frequency.set(text.charAt(i), 1);
			}
		}
		// Check that if given string can be palindrome
		for (let [key, value] of frequency)
		{
			if (value % 2 != 0)
			{
				if (odd == false)
				{
					// Get single character
					odd = true;
				}
				else
				{
					// When have more than two odd character
					status = false;
					break;
				}
			}
		}
		process.stdout.write("\n Given Text : " + text);
		if (status == true)
		{
			// When permutation is form of palindrome
			process.stdout.write("\n Yes ");
			// Auxiliary variables
			var left = "";
			var right = "";
			var middle = "";
			var count = 0;
			// Collect of palindromic sequence
			for (let [key, value] of frequency)
			{
				count = value;
				while (count > 0)
				{
					if (count == 1)
					{
						middle += key;
						count--;
					}
					else
					{
						left = left + key;
						right = key + right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			process.stdout.write(" : [" + left + middle + right + "]");
		}
		else
		{
			process.stdout.write("\n No ");
		}
	}
}

function main()
{
	var task = new Palindrome();
	// Test Cases
	task.isPalindromicPermutation("91002233791");
	task.isPalindromicPermutation("abccbbaa");
	task.isPalindromicPermutation("xyziiyyyx");
}
main();

input

 Given Text : 91002233791
 Yes  : [91023732019]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [xyyiziyyx]
#    Python 3 Program
#    Check if any permutation of string is palindrome
class Palindrome :
	def isPalindromicPermutation(self, text) :
		#  Use to count character frequency
		frequency = dict()
		status = True
		odd = False
		#  Count frequency of given text
		i = 0
		while (i < len(text)) :
			if (text[i] in frequency.keys()) :
				#  Increase character frequency
				frequency[text[i]] = frequency.get(text[i]) + 1
			else :
				#  Add new character with one frequency
				frequency[text[i]] = 1
			
			i += 1
		
		#  Check that if given string can be palindrome
		for key, value in frequency.items() :
			if (value % 2 != 0) :
				if (odd == False) :
					#  Get single character
					odd = True
				else :
					#  When have more than two odd character
					status = False
					break
				
			
		
		print("\n Given Text : ", text, end = "")
		if (status == True) :
			#  When permutation is form of palindrome
			print("\n Yes ", end = "")
			#  Auxiliary variables
			left = ""
			right = ""
			middle = ""
			count = 0
			#  Collect of palindromic sequence
			for key, value in frequency.items() :
				count = value
				while (count > 0) :
					if (count == 1) :
						middle += key
						count -= 1
					else :
						left = left + key
						right = key + right
						count = count - 2
					
				
			
			#  Display a palindrome sequence
			print(" : [", left + middle + right ,"]", end = "")
		else :
			print("\n No ", end = "")
		
	

def main() :
	task = Palindrome()
	#  Test Cases
	task.isPalindromicPermutation("91002233791")
	task.isPalindromicPermutation("abccbbaa")
	task.isPalindromicPermutation("xyziiyyyx")

if __name__ == "__main__": main()

input

 Given Text :  91002233791
 Yes  : [ 09132723190 ]
 Given Text :  abccbbaa
 No
 Given Text :  xyziiyyyx
 Yes  : [ xiyyzyyix ]
#    Ruby Program
#    Check if any permutation of string is palindrome
class Palindrome 
	def isPalindromicPermutation(text) 
		#  Use to count character frequency
		frequency = Hash.new()
		status = true
		odd = false
		#  Count frequency of given text
		i = 0
		while (i < text.length) 
			if (frequency.key?(text[i])) 
				#  Increase character frequency
				frequency[text[i]] = frequency[text[i]] + 1
			else
 
				#  Add new character with one frequency
				frequency[text[i]] = 1
			end

			i += 1
		end

		#  Check that if given string can be palindrome
		frequency.each { | key, value |
			if (value % 2 != 0) 
				if (odd == false) 
					#  Get single character
					odd = true
				else
 
					#  When have more than two odd character
					status = false
					break
				end

			end

		}
		print("\n Given Text : ", text)
		if (status == true) 
			#  When permutation is form of palindrome
			print("\n Yes ")
			#  Auxiliary variables
			left = ""
			right = ""
			middle = ""
			count = 0
			#  Collect of palindromic sequence
			frequency.each { | key, value | 
            count = value
			while (count > 0) 
				if (count == 1) 
					middle += key
					count -= 1
				else
 
					left = left + key
					right = key + right
					count = count - 2
				end

			end
			}
			#  Display a palindrome sequence
			print(" : [", left + middle + right ,"]")
		else
			print("\n No ")
		end

	end

end

def main() 
	task = Palindrome.new()
	#  Test Cases
	task.isPalindromicPermutation("91002233791")
	task.isPalindromicPermutation("abccbbaa")
	task.isPalindromicPermutation("xyziiyyyx")
end

main()

input

 Given Text : 91002233791
 Yes  : [91023732019]
 Given Text : abccbbaa
 No 
 Given Text : xyziiyyyx
 Yes  : [xyyiziyyx]
import scala.collection.mutable._;
import scala.util.control.Breaks._;
/*
    Scala Program
    Check if any permutation of string is palindrome
*/
class Palindrome()
{
	def isPalindromicPermutation(text: String): Unit = {
		// Use to count character frequency
		var frequency: HashMap[Character, Int] = 
          				new HashMap[Character, Int]();
		var status: Boolean = true;
		var odd: Boolean = false;
		// Count frequency of given text
		var i: Int = 0;
		while (i < text.length())
		{
			if (frequency.contains(text.charAt(i)))
			{
				// Increase character frequency
				frequency.addOne(text.charAt(i), 
                                 frequency.get(text.charAt(i)).get + 1);
			}
			else
			{
				// Add new character with one frequency
				frequency.addOne(text.charAt(i), 1);
			}
			i += 1;
		}
		breakable
		{
			// Check that if given string can be palindrome
			for ((key, value) <- frequency)
			{
				if (value % 2 != 0)
				{
					if (odd == false)
					{
						// Get single character
						odd = true;
					}
					else
					{
						// When have more than two odd character
						status = false;
						break;
					}
				}
			}
		}
		print("\n Given Text : " + text);
		if (status == true)
		{
			// When permutation is form of palindrome
			print("\n Yes ");
			// Auxiliary variables
			var left: String = "";
			var right: String = "";
			var middle: String = "";
			var count: Int = 0;
			// Collect of palindromic sequence
			for ((key, value) <-frequency)
			{
				count = value;
				while (count > 0)
				{
					if (count == 1)
					{
						middle += key;
						count -= 1;
					}
					else
					{
						left = left + key;
						right = ""+key + right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			print(" : [" + left + middle + right + "]");
		}
		else
		{
			print("\n No ");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Palindrome = new Palindrome();
		// Test Cases
		task.isPalindromicPermutation("91002233791");
		task.isPalindromicPermutation("abccbbaa");
		task.isPalindromicPermutation("xyziiyyyx");
	}
}

input

 Given Text : 91002233791
 Yes  : [01239793210]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [xiyyzyyix]
import Foundation;
/*
    Swift 4 Program
    Check if any permutation of string is palindrome
*/
class Palindrome
{
	func isPalindromicPermutation(_ t: String)
	{
      	let text = Array(t);
		// Use to count character frequency
		var frequency = [Character : Int]();
		var status = true;
		var odd = false;
		// Count frequency of given text
		var i = 0;
		while (i < text.count)
		{
			if (frequency.keys.contains(text[i]))
			{
				// Increase character frequency
				frequency[text[i]] = frequency[text[i]]! + 1;
			}
			else
			{
				// Add new character with one frequency
				frequency[text[i]] = 1;
			}
			i += 1;
		}
		// Check that if given string can be palindrome
		for (_, value) in frequency
		{
			if (value % 2  != 0)
			{
				if (odd == false)
				{
					// Get single character
					odd = true;
				}
				else
				{
					// When have more than two odd character
					status = false;
					break;
				}
			}
		}
		print("\n Given Text : ", t, terminator: "");
		if (status == true)
		{
			// When permutation is form of palindrome
			print("\n Yes ", terminator: "");
			// Auxiliary variables
			var left = "";
			var right = "";
			var middle = "";
			var count = 0;
			// Collect of palindromic sequence
			for (key, value) in frequency
			{
				count = value;
				while (count > 0)
				{
					if (count == 1)
					{
						middle += String(key);
						count -= 1;
					}
					else
					{
						left = left + String(key);
						right = String(key) + right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			print(" : [", left + middle + right ,"]", terminator: "");
		}
		else
		{
			print("\n No ", terminator: "");
		}
	}
}
func main()
{
	let task = Palindrome();
	// Test Cases
	task.isPalindromicPermutation("91002233791");
	task.isPalindromicPermutation("abccbbaa");
	task.isPalindromicPermutation("xyziiyyyx");
}
main();

input

 Given Text :  91002233791
 Yes  : [ 12390709321 ]
 Given Text :  abccbbaa
 No
 Given Text :  xyziiyyyx
 Yes  : [ xyyiziyyx ]
/*
    Kotlin Program
    Check if any permutation of string is palindrome
*/
class Palindrome
{
	fun isPalindromicPermutation(text: String): Unit
	{
		// Use to count character frequency
		val frequency = mutableMapOf < Char , Int > ();
		var status: Boolean = true;
		var odd: Boolean = false;
		// Count frequency of given text
		var i: Int = 0;
		while (i < text.length)
		{
			if (frequency.containsKey(text.get(i)))
			{
				// Increase character frequency
				frequency.put(text.get(i), frequency.getValue(text.get(i)) + 1);
			}
			else
			{
				// Add new character with one frequency
				frequency.put(text.get(i), 1);
			}
			i += 1;
		}
		// Check that if given string can be palindrome
		for ((_, value) in frequency)
		{
			if (value % 2 != 0)
			{
				if (odd == false)
				{
					// Get single character
					odd = true;
				}
				else
				{
					// When have more than two odd character
					status = false;
					break;
				}
			}
		}
		print("\n Given Text : " + text);
		if (status == true)
		{
			// When permutation is form of palindrome
			print("\n Yes ");
			// Auxiliary variables
			var left: String = "";
			var right: String = "";
			var middle: String = "";
			var count: Int ;
			// Collect of palindromic sequence
			for ((key,value) in frequency)
			{
				count = value;
				while (count > 0)
				{
					if (count == 1)
					{
						middle += key;
						count -= 1;
					}
					else
					{
						left = left + key.toString();
						right = key.toString() + right;
						count = count - 2;
					}
				}
			}
			// Display a palindrome sequence
			print(" : [" + left + middle + right + "]");
		}
		else
		{
			print("\n No ");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Palindrome = Palindrome();
	// Test Cases
	task.isPalindromicPermutation("91002233791");
	task.isPalindromicPermutation("abccbbaa");
	task.isPalindromicPermutation("xyziiyyyx");
}

input

 Given Text : 91002233791
 Yes  : [91023732019]
 Given Text : abccbbaa
 No
 Given Text : xyziiyyyx
 Yes  : [xyyiziyyx]
Last updated on by Kalkicode




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