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Check if K palindromic strings can be formed from a given string

Here given code implementation process.

// Java Program 
// Check if K palindromic strings can be formed from a given string
import java.util.HashMap;

public class Palindrome
{
    // Check whether K palindrome can be generated in character 
    // if rearrangement is allowed in the character
    public boolean constructKPalindrome(String text, int k)
    {

        if( k == text.length()) 
        { 
            // When k is equal to text length
            // Because each character are generate a palindrome
            return true;
        }
        else if(k > text.length())
        {
            // When K is higher of  text length
            return false;
        }
        else
        {
            // Define some auxiliary variables
            int i = 0;
            int count = 0;
            
            // Use to count character frequency
            HashMap < Character, Integer > record = new HashMap < Character, Integer > ();

            // iterate the loop through by length
            // And count character frequency
            for (i = 0; i < text.length(); i++)
            {
           
                if (record.containsKey(text.charAt(i)))
                {
                    record.put(text.charAt(i), record.get(text.charAt(i)) + 1);
                }
                else
                {
                    record.put(text.charAt(i), 1);
                }
            }
            for (int value: record.values())
            {
                if (value % 2 != 0)
                {
                    // When frequency is odd
                   count++;
                }
            }
            if(count > k)
            {
                // When odd frequency character is greater than given K
                return false;
            }
            else 
            {
                return true;
            }
      
        }
    }
    public static void main(String[] args)
    {
        Palindrome task = new Palindrome();

        String text = "aacccaaf";
        // possible palindrome
        int k = 3;
        // text = aacccaaf
        // 1) aa  
        // 2) ccc
        // 3) afa 
      
        System.out.print(" Given text : "+ text);
        System.out.print("\n Given K    : "+ k);

        if(task.constructKPalindrome(text,k))
        {
            System.out.print("\n Result : Possible\n");
        }
        else
        {
            System.out.print("\n Result : Not Possible\n");
        }
    }
}

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
// Include header file
#include <iostream>
#include <string>
#include <unordered_map>
using namespace std;
// C++ Program
// Check if K palindromic strings can be formed from a given string
class Palindrome
{
	public:
		// Check whether K palindrome can be generated in character
		// if rearrangement is allowed in the character
		bool constructKPalindrome(string text, int k)
		{
			if (k == text.length())
			{
				// When k is equal to text length
				// Because each character are generate a palindrome
				return true;
			}
			else if (k > text.length())
			{
				// When K is higher of  text length
				return false;
			}
			else
			{
				// Define some auxiliary variables
				int i = 0;
				int count = 0;
				// Use to count character frequency
				unordered_map < char, int > record;
				// iterate the loop through by length
				// And count character frequency
				for (i = 0; i < text.length(); i++)
				{
					if (record.find(text[i]) != record.end())
					{
						record[text[i]] = record[text[i]] + 1;
					}
					else
					{
						record[text[i]] = 1;
					}
				}
				for (auto &value: record)
				{
					if (value.first % 2 != 0)
					{
						// When frequency is odd
						count++;
					}
				}
				if (count > k)
				{
					// When odd frequency character is greater than given K
					return false;
				}
				else
				{
					return true;
				}
			}
		}
};
int main()
{
	Palindrome task = Palindrome();
	string text = "aacccaaf";
	// possible palindrome
	int k = 3;
	// text = aacccaaf
	// 1) aa
	// 2) ccc
	// 3) afa
	cout << " Given text : " << text;
	cout << "\n Given K    : " << k;
	if (task.constructKPalindrome(text, k))
	{
		cout << "\n Result : Possible\n";
	}
	else
	{
		cout << "\n Result : Not Possible\n";
	}
	return 0;
}

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
// Include namespace system
using System;
using System.Collections.Generic;
// C# Program
// Check if K palindromic strings can be formed from a given string
public class Palindrome
{
	// Check whether K palindrome can be generated in character
	// if rearrangement is allowed in the character
	public Boolean constructKPalindrome(String text, int k)
	{
		if (k == text.Length)
		{
			// When k is equal to text length
			// Because each character are generate a palindrome
			return true;
		}
		else if (k > text.Length)
		{
			// When K is higher of  text length
			return false;
		}
		else
		{
			// Define some auxiliary variables
			int i = 0;
			int count = 0;
			// Use to count character frequency
			Dictionary < char, int > record = new Dictionary < char, int > ();
			// iterate the loop through by length
			// And count character frequency
			for (i = 0; i < text.Length; i++)
			{
				if (record.ContainsKey(text[i]))
				{
					record[text[i]] = record[text[i]] + 1;
				}
				else
				{
					record.Add(text[i], 1);
				}
			}
			foreach(KeyValuePair < char, int > info in record)
			{
				if (info.Value % 2 != 0)
				{
					// When frequency is odd
					count++;
				}
			}
			if (count > k)
			{
				// When odd frequency character is greater than given K
				return false;
			}
			else
			{
				return true;
			}
		}
	}
	public static void Main(String[] args)
	{
		Palindrome task = new Palindrome();
		String text = "aacccaaf";
		// possible palindrome
		int k = 3;
		// text = aacccaaf
		// 1) aa
		// 2) ccc
		// 3) afa
		Console.Write(" Given text : " + text);
		Console.Write("\n Given K    : " + k);
		if (task.constructKPalindrome(text, k))
		{
			Console.Write("\n Result : Possible\n");
		}
		else
		{
			Console.Write("\n Result : Not Possible\n");
		}
	}
}

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
<?php
// Php Program
// Check if K palindromic strings can be formed from a given string
class Palindrome
{
	// Check whether K palindrome can be generated in character
	// if rearrangement is allowed in the character
	public	function constructKPalindrome($text, $k)
	{
		if ($k == strlen($text))
		{
			// When k is equal to text length
			// Because each character are generate a palindrome
			return true;
		}
		else if ($k > strlen($text))
		{
			// When K is higher of  text length
			return false;
		}
		else
		{
			// Define some auxiliary variables
			$i = 0;
			$count = 0;
			// Use to count character frequency
			$record = array();
			// iterate the loop through by length
			// And count character frequency
			for ($i = 0; $i < strlen($text); $i++)
			{
				if (array_key_exists($text[$i], $record))
				{
					$record[$text[$i]] = $record[$text[$i]] + 1;
				}
				else
				{
					$record[$text[$i]] = 1;
				}
			}
			foreach($record as $key => $value)
			{
				if ($value % 2 != 0)
				{
					// When frequency is odd
					$count++;
				}
			}
			if ($count > $k)
			{
				// When odd frequency character is greater than given K
				return false;
			}
			else
			{
				return true;
			}
		}
	}
}

function main()
{
	$task = new Palindrome();
	$text = "aacccaaf";
	// possible palindrome
	$k = 3;
	// text = aacccaaf
	// 1) aa
	// 2) ccc
	// 3) afa
	echo " Given text : ". $text;
	echo "\n Given K    : ". $k;
	if ($task->constructKPalindrome($text, $k))
	{
		echo "\n Result : Possible\n";
	}
	else
	{
		echo "\n Result : Not Possible\n";
	}
}
main();

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
// Node Js Program
// Check if K palindromic strings can be formed from a given string
class Palindrome
{
	// Check whether K palindrome can be generated in character
	// if rearrangement is allowed in the character
	constructKPalindrome(text, k)
	{
		if (k == text.length)
		{
			// When k is equal to text length
			// Because each character are generate a palindrome
			return true;
		}
		else if (k > text.length)
		{
			// When K is higher of  text length
			return false;
		}
		else
		{
			// Define some auxiliary variables
			var i = 0;
			var count = 0;
			// Use to count character frequency
			var record = new Map();
			// iterate the loop through by length
			// And count character frequency
			for (i = 0; i < text.length; i++)
			{
				if (record.has(text.charAt(i)))
				{
					record.set(text.charAt(i), record.get(text.charAt(i)) + 1);
				}
				else
				{
					record.set(text.charAt(i), 1);
				}
			}
			for (let [key, value] of record)
			{
				if (value % 2 != 0)
				{
					// When frequency is odd
					count++;
				}
			}
			if (count > k)
			{
				// When odd frequency character is greater than given K
				return false;
			}
			else
			{
				return true;
			}
		}
	}
}

function main()
{
	var task = new Palindrome();
	var text = "aacccaaf";
	// possible palindrome
	var k = 3;
	// text = aacccaaf
	// 1) aa
	// 2) ccc
	// 3) afa
	process.stdout.write(" Given text : " + text);
	process.stdout.write("\n Given K    : " + k);
	if (task.constructKPalindrome(text, k))
	{
		process.stdout.write("\n Result : Possible\n");
	}
	else
	{
		process.stdout.write("\n Result : Not Possible\n");
	}
}
main();

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
#  Python 3 Program
#  Check if K palindromic strings can be formed from a given string
class Palindrome :
	#  Check whether K palindrome can be generated in character
	#  if rearrangement is allowed in the character
	def constructKPalindrome(self, text, k) :
		if (k == len(text)) :
			#  When k is equal to text length
			#  Because each character are generate a palindrome
			return True
		elif(k > len(text)) :
			#  When K is higher of  text length
			return False
		else :
			#  Define some auxiliary variables
			i = 0
			count = 0
			#  Use to count character frequency
			record = dict()
			#  iterate the loop through by length
			#  And count character frequency
			while (i < len(text)) :
				if (text[i] in record.keys()) :
					record[text[i]] = record.get(text[i]) + 1
				else :
					record[text[i]] = 1
				i += 1
			
			for key, value in record.items() :
				if (value % 2 != 0) :
					#  When frequency is odd
					count += 1
				
			
			if (count > k) :
				#  When odd frequency character is greater than given K
				return False
			else :
				return True
			

def main() :
	task = Palindrome()
	text = "aacccaaf"
	#  possible palindrome
	k = 3
	#  text = aacccaaf
	#  1) aa
	#  2) ccc
	#  3) afa
	print(" Given text : ", text, end = "")
	print("\n Given K    : ", k, end = "")
	if (task.constructKPalindrome(text, k)) :
		print("\n Result : Possible")
	else :
		print("\n Result : Not Possible")
	

if __name__ == "__main__": main()

Output

 Given text :  aacccaaf
 Given K    :  3
 Result : Possible
#  Ruby Program
#  Check if K palindromic strings can be formed from a given string
class Palindrome 
	#  Check whether K palindrome can be generated in character
	#  if rearrangement is allowed in the character
	def constructKPalindrome(text, k) 
		if (k == text.length) 
			#  When k is equal to text length
			#  Because each character are generate a palindrome
			return true
		elsif(k > text.length) 
			#  When K is higher of  text length
			return false
		else 
			#  Define some auxiliary variables
			i = 0
			count = 0
			#  Use to count character frequency
			record = Hash.new
			#  iterate the loop through by length
			#  And count character frequency
			while (i < text.length) 
				if (record.key?(text[i])) 
					record[text[i]] = record[text[i]] + 1
				else 
					record[text[i]] = 1
				end

				i += 1
			end

			record.each { | key, value | 
				if (value % 2 != 0) 
					#  When frequency is odd
					count += 1
				end
			}
			if (count > k) 
				#  When odd frequency character is greater than given K
				return false
			else 
				return true
			end

		end

	end

end

def main() 
	task = Palindrome.new()
	text = "aacccaaf"
	#  possible palindrome
	k = 3
	#  text = aacccaaf
	#  1) aa
	#  2) ccc
	#  3) afa
	print(" Given text : ", text)
	print("\n Given K    : ", k)
	if (task.constructKPalindrome(text, k)) 
		print("\n Result : Possible\n")
	else 
		print("\n Result : Not Possible\n")
	end

end

main()

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible

import scala.collection.mutable._;
// Scala Program
// Check if K palindromic strings can be formed from a given string
class Palindrome
{
	// Check whether K palindrome can be generated in character
	// if rearrangement is allowed in the character
	def constructKPalindrome(text: String, k: Int): Boolean = {
		if (k == text.length())
		{
			// When k is equal to text length
			// Because each character are generate a palindrome
			return true;
		}
		else if (k > text.length())
		{
			// When K is higher of  text length
			return false;
		}
		else
		{
			// Define some auxiliary variables
			var i: Int = 0;
			var count: Int = 0;
			// Use to count character frequency
			var record = Map[Char, Int]();
			// iterate the loop through by length
			// And count character frequency
			while (i < text.length())
			{
				if (record.contains(text.charAt(i)))
				{
					record.addOne(text.charAt(i), 
                                  record.get(text.charAt(i)).get + 1);
				}
				else
				{
					record.addOne(text.charAt(i), 1);
				}
				i += 1;
			}
			for ((key, v) <-  record)
			{
				if (v % 2 != 0)
				{
					// When frequency is odd
					count += 1;
				}
			}
			if (count > k)
			{
				// When odd frequency character is greater than given K
				return false;
			}
			else
			{
				return true;
			}
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Palindrome = new Palindrome();
		var text: String = "aacccaaf";
		// possible palindrome
		var k: Int = 3;
		// text = aacccaaf
		// 1) aa
		// 2) ccc
		// 3) afa
		print(" Given text : " + text);
		print("\n Given K    : " + k);
		if (task.constructKPalindrome(text, k))
		{
			print("\n Result : Possible\n");
		}
		else
		{
			print("\n Result : Not Possible\n");
		}
	}
}

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
import Foundation
// Swift 4 Program
// Check if K palindromic strings can be formed from a given string
class Palindrome
{
	// Check whether K palindrome can be generated in character
	// if rearrangement is allowed in the character
	func constructKPalindrome(_ t: String, _ k: Int)->Bool
	{
      	var text = Array(t);
		if (k == text.count)
		{
			// When k is equal to text length
			// Because each character are generate a palindrome
			return true;
		}
		else if (k > text.count)
		{
			// When K is higher of  text length
			return false;
		}
		else
		{
			// Define some auxiliary variables
			var i: Int = 0;
			var count: Int = 0;
			// Use to count character frequency
			var record = [Character: Int]();
			// iterate the loop through by length
			// And count character frequency
			while (i < text.count)
			{
				if (record.keys.contains(text[i]))
				{
					record[text[i]] = record[text[i]]! + 1;
				}
				else
				{
					record[text[i]] = 1;
				}
				i += 1;
			}
			for (_, value) in record
			{
				if (value % 2  != 0)
				{
					// When frequency is odd
					count += 1;
				}
			}
			if (count > k)
			{
				// When odd frequency character is greater than given K
				return false;
			}
			else
			{
				return true;
			}
		}
	}
}
func main()
{
	let task: Palindrome = Palindrome();
	let text: String = "aacccaaf";
	// possible palindrome
	let k: Int = 3;
	// text = aacccaaf
	// 1) aa
	// 2) ccc
	// 3) afa
	print(" Given text : ", text, terminator: "");
	print("\n Given K    : ", k, terminator: "");
	if (task.constructKPalindrome(text, k))
	{
		print("\n Result : Possible");
	}
	else
	{
		print("\n Result : Not Possible");
	}
}
main();

Output

 Given text :  aacccaaf
 Given K    :  3
 Result : Possible
// Kotlin Program
// Check if K palindromic strings can be formed from a given string
class Palindrome
{
	// Check whether K palindrome can be generated in character
	// if rearrangement is allowed in the character
	fun constructKPalindrome(text: String, k: Int): Boolean
	{
		if (k == text.length)
		{
			// When k is equal to text length
			// Because each character are generate a palindrome
			return true;
		}
		else if (k > text.length)
		{
			// When K is higher of  text length
			return false;
		}
		else
		{
			// Define some auxiliary variables
			var i: Int = 0;
			var count: Int = 0;
			// Use to count character frequency
			var record = mutableMapOf < Char , Int > ();
			// iterate the loop through by length
			// And count character frequency
			while (i < text.length)
			{
				if (record.containsKey(text.get(i)))
				{
					record.put(text.get(i), record.getValue(text.get(i)) + 1);
				}
				else
				{
					record.put(text.get(i), 1);
				}
				i += 1;
			}
			for (value in record.values)
			{
				if (value % 2 != 0)
				{
					// When frequency is odd
					count += 1;
				}
			}
			if (count > k)
			{
				// When odd frequency character is greater than given K
				return false;
			}
			else
			{
				return true;
			}
		}
	}
}
fun main(args: Array < String > ): Unit
{
	var task: Palindrome = Palindrome();
	var text: String = "aacccaaf";
	// possible palindrome
	var k: Int = 3;
	// text = aacccaaf
	// 1) aa
	// 2) ccc
	// 3) afa
	print(" Given text : " + text);
	print("\n Given K    : " + k);
	if (task.constructKPalindrome(text, k))
	{
		print("\n Result : Possible\n");
	}
	else
	{
		print("\n Result : Not Possible\n");
	}
}

Output

 Given text : aacccaaf
 Given K    : 3
 Result : Possible
Last updated on by Kalkicode




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