Posted on by Kalkicode
Code Hash

Minimum number of deletion required to make two strings anagram

Here given code implementation process.

// Include header file
#include <iostream>
#include <string>
#include <unordered_map>

using namespace std;
/*
  C++ program for
  Minimum number of deletion required to make two strings anagram
*/
class Anagrams
{
    public: void countFrequency(string str, unordered_map < char, int > &record)
    {
        int n = str.length();
        // Calculate frequency of given string
        for (int i = 0; i < n; ++i)
        {
            if (record.find(str[i]) != record.end())
            {
                record[str[i]] = record[str[i]] + 1;
            }
            else
            {
                record[str[i]] = 1;
            }
        }
    }
    void makeAnagram(string str1, string str2)
    {
        int operation = 0;
        unordered_map < char, int > record1;
        unordered_map < char, int > record2;
        // Count frequency of given string
        this->countFrequency(str1, record1);
        this->countFrequency(str2, record2);
        int k1 = 0;
        int k2 = 0;
        // Compare same frequency in both string
        for (auto &info: record1)
        {
            if (record2.find(info.first) != record2.end())
            {
                k1 = info.second;
                k2 = record2[info.first];
                if (k1 > k2)
                {
                    operation += k1 - k2;
                }
                else if (k2 > k1)
                {
                    operation += k2 - k1;
                }
                // Reset frequency value
                record1[info.first] = 0;
                record2[info.first] = 0;
            }
        }
        // Add number of extra characters in str1
        for (auto &info: record1)
        {
            operation += info.second;
        }
        // Add number of extra characters in str2
        for (auto &info: record2)
        {
            operation += info.second;
        }
        cout << " Given (" << str1 << "),(" << str2 << ")" << endl;
        // Display number of remove operation to make anagram
        cout << " Result : " << operation << endl;
    }
};
int main()
{
    Anagrams *task = new Anagrams();
    /*
        Case A
        ------
        k m e t e o r o i l , r e m f i l l o t e
        ↓             ↓             ↓     ↓
        ①            ②             ③    ④
        ------------------------------------------------
        Remove 4 characters to make both string is anagram
    */
    task->makeAnagram("kmeteoroil", "remfillote");
    /*
        Case B
        ------
        C  S I T W I  N  , W  S  O  I  C  N  L  P
               ↓   ↓             ↓           ↓  ↓
               ①  ②             ③          ④  ⑤
        ----------------------------------------------
        Remove 5 characters to make both string is anagram
    */
    task->makeAnagram("CSITWIN", "WSOICNLP");
    /*
        Case C
        ------
        Both string is anagram
    */
    task->makeAnagram("123", "321");
    return 0;
}

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
import java.util.HashMap;
/*
  Java program for
  Minimum number of deletion required to make two strings anagram
*/
public class Anagrams
{
	public void countFrequency(String str, HashMap < Character, Integer > record)
	{
		int n = str.length();
		// Calculate frequency of given string
		for (int i = 0; i < n; ++i)
		{
			if (record.containsKey(str.charAt(i)))
			{
				record.put(str.charAt(i), record.get(str.charAt(i)) + 1);
			}
			else
			{
				record.put(str.charAt(i), 1);
			}
		}
	}
	public void makeAnagram(String str1, String str2)
	{
		int operation = 0;
		HashMap < Character, Integer > record1 = new HashMap < Character, Integer > ();
		HashMap < Character, Integer > record2 = new HashMap < Character, Integer > ();
		// Count frequency of given string
		countFrequency(str1, record1);
		countFrequency(str2, record2);
		int k1 = 0;
		int k2 = 0;
		// Compare same frequency in both string
		for (char info: record1.keySet())
		{
			if (record2.containsKey(info))
			{
				k1 = record1.get(info);
				k2 = record2.get(info);
				if (k1 > k2)
				{
					operation += k1 - k2;
				}
				else if (k2 > k1)
				{
					operation += k2 - k1;
				}
				// Reset frequency value
				record1.put(info, 0);
				record2.put(info, 0);
			}
		}
		// Add number of extra characters in str1
		for (char info: record1.keySet())
		{
			operation += record1.get(info);
		}
		// Add number of extra characters in str2
		for (char info: record2.keySet())
		{
			operation += record2.get(info);
		}
		System.out.println(" Given (" + str1 + "),(" + str2 + ")");
		// Display number of remove operation to make anagram
		System.out.println(" Result : " + operation);
	}
	public static void main(String[] args)
	{
		Anagrams task = new Anagrams();
		/*
		    Case A
		    ------
		    k m e t e o r o i l , r e m f i l l o t e
		    ↓             ↓             ↓     ↓
		    ①            ②             ③    ④
		    ------------------------------------------------
		    Remove 4 characters to make both string is anagram
		*/
		task.makeAnagram("kmeteoroil", "remfillote");
		/*
		    Case B
		    ------
		    C  S I T W I  N  , W  S  O  I  C  N  L  P
		           ↓   ↓             ↓           ↓  ↓
		           ①  ②             ③          ④  ⑤
		    ----------------------------------------------
		    Remove 5 characters to make both string is anagram
		*/
		task.makeAnagram("CSITWIN", "WSOICNLP");
		/*
		    Case C
		    ------
		    Both string is anagram
		*/
		task.makeAnagram("123", "321");
	}
}

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
// Include namespace system
using System;
using System.Collections.Generic;
/*
  Csharp program for
  Minimum number of deletion required to make two strings anagram
*/
public class Anagrams
{
	public void countFrequency(String str, Dictionary < char, int > record)
	{
		int n = str.Length;
		// Calculate frequency of given string
		for (int i = 0; i < n; ++i)
		{
			if (record.ContainsKey(str[i]))
			{
				record[str[i]] = record[str[i]] + 1;
			}
			else
			{
				record.Add(str[i], 1);
			}
		}
	}
	public void makeAnagram(String str1, String str2)
	{
		int operation = 0;
		Dictionary < char, int > record1 = 
          new Dictionary < char, int > ();
		Dictionary < char, int > record2 = 
          new Dictionary < char, int > ();
		// Count frequency of given string
		this.countFrequency(str1, record1);
		this.countFrequency(str2, record2);
		int k1 = 0;
		int k2 = 0;
      	List<char> keys = new List<char>(record1.Keys);
		// Compare same frequency in both string
		foreach(char key in keys)
		{
			if (record2.ContainsKey(key))
			{
				k1 = record1[key];
				k2 = record2[key];
				if (k1 > k2)
				{
					operation += k1 - k2;
				}
				else if (k2 > k1)
				{
					operation += k2 - k1;
				}
				// Reset frequency value
				record1[key] = 0;
				record2[key] = 0;
			}
		}
		// Add number of extra characters in str1
		foreach(KeyValuePair < char, int > info in record1)
		{
			operation += info.Value;
		}
		// Add number of extra characters in str2
		foreach(KeyValuePair < char, int > info in record2)
		{
			operation += info.Value;
		}
		Console.WriteLine(" Given (" + str1 + "),(" + str2 + ")");
		// Display number of remove operation to make anagram
		Console.WriteLine(" Result : " + operation);
	}
	public static void Main(String[] args)
	{
		Anagrams task = new Anagrams();
		/*
		    Case A
		    ------
		    k m e t e o r o i l , r e m f i l l o t e
		    ↓             ↓             ↓     ↓
		    ①            ②             ③    ④
		    ------------------------------------------------
		    Remove 4 characters to make both string is anagram
		*/
		task.makeAnagram("kmeteoroil", "remfillote");
		/*
		    Case B
		    ------
		    C  S I T W I  N  , W  S  O  I  C  N  L  P
		           ↓   ↓             ↓           ↓  ↓
		           ①  ②             ③          ④  ⑤
		    ----------------------------------------------
		    Remove 5 characters to make both string is anagram
		*/
		task.makeAnagram("CSITWIN", "WSOICNLP");
		/*
		    Case C
		    ------
		    Both string is anagram
		*/
		task.makeAnagram("123", "321");
	}
}

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
package main
import "fmt"
/*
  Go program for
  Minimum number of deletion required to make two strings anagram
*/

func countFrequency(str string, record map[byte]int) {
	var n int = len(str)
	// Calculate frequency of given string
	for i := 0 ; i < n ; i++ {
		if _, found := record[str[i]] ; found {
			record[str[i]] = record[str[i]] + 1
		} else {
			record[str[i]] = 1
		}
	}
}
func makeAnagram(str1, str2 string) {
	var operation int = 0
	var record1 = make(map[byte] int)
	var record2 = make(map[byte] int)
	// Count frequency of given string
	countFrequency(str1, record1)
	countFrequency(str2, record2)
	var k1 int = 0
	var k2 int = 0
	// Compare same frequency in both string
	for k, v := range record1 {
		if _, found := record2[k] ; found {
			k1 = v
			k2 = record2[k]
			if k1 > k2 {
				operation += k1 - k2
			} else if k2 > k1 {
				operation += k2 - k1
			}
			// Reset frequency value
			record1[k] = 0
			record2[k] = 0
		}
	}
	// Add number of extra characters in str1
	for _, v := range record1 {
		operation += v
	}
	// Add number of extra characters in str2
	for _, v := range record2 {
		operation += v
	}
	fmt.Println(" Given (", str1, "),(", str2, ")")
	// Display number of remove operation to make anagram
	fmt.Println(" Result : ", operation)
}
func main() {

	/*
	    Case A
	    ------
	    k m e t e o r o i l , r e m f i l l o t e
	    ↓             ↓             ↓     ↓
	    ①            ②             ③    ④
	    ------------------------------------------------
	    Remove 4 characters to make both string is anagram
	*/
	makeAnagram("kmeteoroil", "remfillote")
	/*
	    Case B
	    ------
	    C  S I T W I  N  , W  S  O  I  C  N  L  P
	           ↓   ↓             ↓           ↓  ↓
	           ①  ②             ③          ④  ⑤
	    ----------------------------------------------
	    Remove 5 characters to make both string is anagram
	*/
	makeAnagram("CSITWIN", "WSOICNLP")
	/*
	    Case C
	    ------
	    Both string is anagram
	*/
	makeAnagram("123", "321")
}

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
<?php
/*
  Php program for
  Minimum number of deletion required to make two strings anagram
*/
class Anagrams
{
	public	function countFrequency($str, &$record)
	{
		$n = strlen($str);
		// Calculate frequency of given string
		for ($i = 0; $i < $n; ++$i)
		{
			if (array_key_exists($str[$i], $record))
			{
				$record[$str[$i]] = $record[$str[$i]] + 1;
			}
			else
			{
				$record[$str[$i]] = 1;
			}
		}
	}
	public	function makeAnagram($str1, $str2)
	{
		$operation = 0;
		$record1 = array();
		$record2 = array();
		// Count frequency of given string
		$this->countFrequency($str1, $record1);
		$this->countFrequency($str2, $record2);
		$k1 = 0;
		$k2 = 0;
		// Compare same frequency in both string
		foreach($record1 as $key => $value)
		{
			if (array_key_exists($key, $record2))
			{
				$k1 = $record1[$key];
				$k2 = $record2[$key];
				if ($k1 > $k2)
				{
					$operation += $k1 - $k2;
				}
				else if ($k2 > $k1)
				{
					$operation += $k2 - $k1;
				}
				// Reset frequency value
				$record1[$key] = 0;
				$record2[$key] = 0;
			}
		}
		// Add number of extra characters in str1
		foreach($record1 as $key => $value)
		{
			$operation += $value;
		}
		// Add number of extra characters in str2
		foreach($record2 as $key => $value)
		{
			$operation += $value;
		}
		echo(" Given (".$str1."),(".$str2.")\n");
		// Display number of remove operation to make anagram
		echo(" Result : ".$operation."\n");
	}
}

function main()
{
	$task = new Anagrams();
	/*
	    Case A
	    ------
	    k m e t e o r o i l , r e m f i l l o t e
	    ↓             ↓             ↓     ↓
	    ①            ②             ③    ④
	    ------------------------------------------------
	    Remove 4 characters to make both string is anagram
	*/
	$task->makeAnagram("kmeteoroil", "remfillote");
	/*
	    Case B
	    ------
	    C  S I T W I  N  , W  S  O  I  C  N  L  P
	           ↓   ↓             ↓           ↓  ↓
	           ①  ②             ③          ④  ⑤
	    ----------------------------------------------
	    Remove 5 characters to make both string is anagram
	*/
	$task->makeAnagram("CSITWIN", "WSOICNLP");
	/*
	    Case C
	    ------
	    Both string is anagram
	*/
	$task->makeAnagram("123", "321");
}
main();

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
/*
  Node JS program for
  Minimum number of deletion required to make two strings anagram
*/
class Anagrams
{
	countFrequency(str, record)
	{
		var n = str.length;
		// Calculate frequency of given string
		for (var i = 0; i < n; ++i)
		{
			if (record.has(str.charAt(i)))
			{
				record.set(str.charAt(i), record.get(str.charAt(i)) + 1);
			}
			else
			{
				record.set(str.charAt(i), 1);
			}
		}
	}
	makeAnagram(str1, str2)
	{
		var operation = 0;
		var record1 = new Map();
		var record2 = new Map();
		// Count frequency of given string
		this.countFrequency(str1, record1);
		this.countFrequency(str2, record2);
		var k1 = 0;
		var k2 = 0;
		// Compare same frequency in both string
		for (let [key, value] of record1)
		{
			if (record2.has(key))
			{
				k1 = value;
				k2 = record2.get(key);
				if (k1 > k2)
				{
					operation += k1 - k2;
				}
				else if (k2 > k1)
				{
					operation += k2 - k1;
				}
				// Reset frequency value
				record1.set(key, 0);
				record2.set(key, 0);
			}
		}
		// Add number of extra characters in str1
		for (let [key, value] of record1)
		{
			operation += value;
		}
		// Add number of extra characters in str2
		for (let [key, value] of record2)
		{
			operation += value;
		}
		console.log(" Given (" + str1 + "),(" + str2 + ")");
		// Display number of remove operation to make anagram
		console.log(" Result : " + operation);
	}
}

function main()
{
	var task = new Anagrams();
	/*
	    Case A
	    ------
	    k m e t e o r o i l , r e m f i l l o t e
	    ↓             ↓             ↓     ↓
	    ①            ②             ③    ④
	    ------------------------------------------------
	    Remove 4 characters to make both string is anagram
	*/
	task.makeAnagram("kmeteoroil", "remfillote");
	/*
	    Case B
	    ------
	    C  S I T W I  N  , W  S  O  I  C  N  L  P
	           ↓   ↓             ↓           ↓  ↓
	           ①  ②             ③          ④  ⑤
	    ----------------------------------------------
	    Remove 5 characters to make both string is anagram
	*/
	task.makeAnagram("CSITWIN", "WSOICNLP");
	/*
	    Case C
	    ------
	    Both string is anagram
	*/
	task.makeAnagram("123", "321");
}
main();

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
#  Python 3 program for
#  Minimum number of deletion required to make two strings anagram
class Anagrams :
	def countFrequency(self, str, record) :
		n = len(str)
		i = 0
		#  Calculate frequency of given string
		while (i < n) :
			if ((str[i] in record.keys())) :
				record[str[i]] = record.get(str[i]) + 1
			else :
				record[str[i]] = 1
			
			i += 1
		
	
	def makeAnagram(self, str1, str2) :
		operation = 0
		record1 = dict()
		record2 = dict()
		#  Count frequency of given string
		self.countFrequency(str1, record1)
		self.countFrequency(str2, record2)
		k1 = 0
		k2 = 0
		for key, value in record1.items() :
			if ((key in record2.keys())) :
				k1 = value
				k2 = record2.get(key)
				if (k1 > k2) :
					operation += k1 - k2
				elif (k2 > k1) :
					operation += k2 - k1
				
				#  Reset frequency value
				record1[key] = 0
				record2[key] = 0
			
		
		for key, value in record1.items() :
			operation += value
		
		for key, value in record2.items() :
			operation += value
		
		print(" Given (", str1 ,"),(", str2 ,")",sep="")
		#  Display number of remove operation to make anagram
		print(" Result : ", operation)
	

def main() :
	task = Anagrams()
	#    Case A
	#    ------
	#    k m e t e o r o i l , r e m f i l l o t e
	#    ↓             ↓             ↓     ↓
	#    ①            ②             ③    ④
	#    ------------------------------------------------
	#    Remove 4 characters to make both string is anagram
	task.makeAnagram("kmeteoroil", "remfillote")
	#    Case B
	#    ------
	#    C  S I T W I  N  , W  S  O  I  C  N  L  P
	#           ↓   ↓             ↓           ↓  ↓
	#           ①  ②             ③          ④  ⑤
	#    ----------------------------------------------
	#    Remove 5 characters to make both string is anagram
	task.makeAnagram("CSITWIN", "WSOICNLP")
	#    Case C
	#    ------
	#    Both string is anagram
	task.makeAnagram("123", "321")

if __name__ == "__main__": main()

Output

 Given (kmeteoroil),(remfillote)
 Result :  4
 Given (CSITWIN),(WSOICNLP)
 Result :  5
 Given (123),(321)
 Result :  0
#  Ruby program for
#  Minimum number of deletion required to make two strings anagram
class Anagrams 
	def countFrequency(str, record) 
		n = str.length
		i = 0
		#  Calculate frequency of given string
		while (i < n) 
			if (record.key?(str[i])) 
				record[str[i]] = record[str[i]] + 1
			else
				record[str[i]] = 1
			end
			i += 1
		end

	end

	def makeAnagram(str1, str2) 
		operation = 0
		record1 = Hash.new()
		record2 = Hash.new()
		#  Count frequency of given string
		self.countFrequency(str1, record1)
		self.countFrequency(str2, record2)
		k1 = 0
		k2 = 0
		#  Compare same frequency in both string
		record1.each { | key, value |
			if (record2.key?(key)) 
				k1 = record1[key]
				k2 = record2[key]
				if (k1 > k2) 
					operation += k1 - k2
				elsif (k2 > k1) 
					operation += k2 - k1
				end

				#  Reset frequency value
				record1[key] = 0
				record2[key] = 0
			end

		}
		#  Add number of extra characters in str1
		record1.each { | key, value | operation += value}
		#  Add number of extra characters in str2
		record2.each { | key, value | operation += value }
		print(" Given (", str1 ,"),(", str2 ,")", "\n")
		#  Display number of remove operation to make anagram
		print(" Result : ", operation, "\n")
	end

end

def main() 
	task = Anagrams.new()
	#    Case A
	#    ------
	#    k m e t e o r o i l , r e m f i l l o t e
	#    ↓             ↓             ↓     ↓
	#    ①            ②             ③    ④
	#    ------------------------------------------------
	#    Remove 4 characters to make both string is anagram
	task.makeAnagram("kmeteoroil", "remfillote")
	#    Case B
	#    ------
	#    C  S I T W I  N  , W  S  O  I  C  N  L  P
	#           ↓   ↓             ↓           ↓  ↓
	#           ①  ②             ③          ④  ⑤
	#    ----------------------------------------------
	#    Remove 5 characters to make both string is anagram
	task.makeAnagram("CSITWIN", "WSOICNLP")
	#    Case C
	#    ------
	#    Both string is anagram
	task.makeAnagram("123", "321")
end

main()

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0

import scala.collection.mutable._;
/*
  Scala program for
  Minimum number of deletion required to make two strings anagram
*/
class Anagrams()
{
	def countFrequency(str: String, 
                       record: Map[Character, Int]): Unit = {
		var n: Int = str.length();
		var i: Int = 0;
		// Calculate frequency of given string
		while (i < n)
		{
			if (record.contains(str.charAt(i)))
			{
				record.addOne(str.charAt(i), 
                              record.get(str.charAt(i)).get + 1);
			}
			else
			{
				record.addOne(str.charAt(i), 1);
			}
			i += 1;
		}
	}
	def makeAnagram(str1: String, str2: String): Unit = {
		var operation: Int = 0;
		var record1: Map[Character, Int] = Map();
		var record2: Map[Character, Int] = Map();
		// Count frequency of given string
		countFrequency(str1, record1);
		countFrequency(str2, record2);
		var k1: Int = 0;
		var k2: Int = 0;
		// Compare same frequency in both string
		
		for ((key,value) <- record1)
 		{
			if (record2.contains(key))
			{
				k1 = value;
				k2 = record2.get(key).get;
				if (k1 > k2)
				{
					operation += k1 - k2;
				}
				else if (k2 > k1)
				{
					operation += k2 - k1;
				}
				// Reset frequency value
				record1.addOne(key, 0);
				record2.addOne(key, 0);
			}
		}
		// Add number of extra characters in str1
		for ((key,value) <- record1)
		{
			operation += value;
		}
		// Add number of extra characters in str2
		for ((key,value) <- record2)
		{
			operation += value;
		}
		println(" Given (" + str1 + "),(" + str2 + ")");
		// Display number of remove operation to make anagram
		println(" Result : " + operation);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Anagrams = new Anagrams();
		/*
		    Case A
		    ------
		    k m e t e o r o i l , r e m f i l l o t e
		    ↓             ↓             ↓     ↓
		    ①            ②             ③    ④
		    ------------------------------------------------
		    Remove 4 characters to make both string is anagram
		*/
		task.makeAnagram("kmeteoroil", "remfillote");
		/*
		    Case B
		    ------
		    C  S I T W I  N  , W  S  O  I  C  N  L  P
		           ↓   ↓             ↓           ↓  ↓
		           ①  ②             ③          ④  ⑤
		    ----------------------------------------------
		    Remove 5 characters to make both string is anagram
		*/
		task.makeAnagram("CSITWIN", "WSOICNLP");
		/*
		    Case C
		    ------
		    Both string is anagram
		*/
		task.makeAnagram("123", "321");
	}
}

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0
import Foundation;
/*
  Swift 4 program for
  Minimum number of deletion required to make two strings anagram
*/
class Anagrams
{
	func countFrequency(_ str: [Character], _ record: inout[Character:Int])
	{
		let n: Int = str.count;
		var i: Int = 0;
		// Calculate frequency of given string
		while (i < n)
		{
			if (record.keys.contains(str[i]))
			{
				record[str[i]] = record[str[i]]! + 1;
			}
			else
			{
				record[str[i]] = 1;
			}
			i += 1;
		}
	}
	func makeAnagram(_ str1: String, _ str2: String)
	{
		var operation: Int = 0;
		var record1 = [Character : Int]();
		var record2 = [Character : Int]();
		// Count frequency of given string
		self.countFrequency(Array(str1), &record1);
		self.countFrequency(Array(str2), &record2);
		var k1: Int = 0;
		var k2: Int = 0;
		// Compare same frequency in both string
		for (key, value) in record1
		{
			if (record2.keys.contains(key))
			{
				k1 = value;
				k2 = record2[key]!;
				if (k1 > k2)
				{
					operation += k1 - k2;
				}
				else if (k2 > k1)
				{
					operation += k2 - k1;
				}
				// Reset frequency value
				record1[key] = 0;
				record2[key] = 0;
			}
		}
		// Add number of extra characters in str1
		for (_, value) in record1
		{
			operation += value;
		}
		// Add number of extra characters in str2
		for (_, value) in record2
		{
			operation += value;
		}
		print(" Given (", str1 ,"),(", str2 ,")");
		// Display number of remove operation to make anagram
		print(" Result : ", operation);
	}
}
func main()
{
	let task: Anagrams = Anagrams();
	/*
	    Case A
	    ------
	    k m e t e o r o i l , r e m f i l l o t e
	    ↓             ↓             ↓     ↓
	    ①            ②             ③    ④
	    ------------------------------------------------
	    Remove 4 characters to make both string is anagram
	*/
	task.makeAnagram("kmeteoroil", "remfillote");
	/*
	    Case B
	    ------
	    C  S I T W I  N  , W  S  O  I  C  N  L  P
	           ↓   ↓             ↓           ↓  ↓
	           ①  ②             ③          ④  ⑤
	    ----------------------------------------------
	    Remove 5 characters to make both string is anagram
	*/
	task.makeAnagram("CSITWIN", "WSOICNLP");
	/*
	    Case C
	    ------
	    Both string is anagram
	*/
	task.makeAnagram("123", "321");
}
main();

Output

 Given ( kmeteoroil ),( remfillote )
 Result :  4
 Given ( CSITWIN ),( WSOICNLP )
 Result :  5
 Given ( 123 ),( 321 )
 Result :  0
/*
  Kotlin program for
  Minimum number of deletion required to make two strings anagram
*/
class Anagrams
{
	fun countFrequency(str: String, 
                        record: MutableMap <Char, Int > ): Unit
	{
		val n: Int = str.length;
		var i: Int = 0;
		// Calculate frequency of given string
		while (i < n)
		{
			if (record.containsKey(str.get(i)))
			{
				record.put(str.get(i), record.getValue(str.get(i)) + 1);
			}
			else
			{
				record.put(str.get(i), 1);
			}
			i += 1;
		}
	}
	fun makeAnagram(str1: String, str2: String): Unit
	{
		var operation: Int = 0;
		var record1 = mutableMapOf<Char, Int>();
		var record2 = mutableMapOf<Char, Int>();
		// Count frequency of given string
		this.countFrequency(str1, record1);
		this.countFrequency(str2, record2);
		var k1: Int ;
		var k2: Int ;
		// Compare same frequency in both string
		for ((key, value) in record1)
		{
			if (record2.containsKey(key))
			{
				k1 = value;
				k2 = record2.getValue(key);
				if (k1 > k2)
				{
					operation += k1 - k2;
				}
				else if (k2 > k1)
				{
					operation += k2 - k1;
				}
				// Reset frequency value
				record1.put(key, 0);
				record2.put(key, 0);
			}
		}
		// Add number of extra characters in str1
		for ((_, value) in record1)
		{
			operation += value;
		}
		// Add number of extra characters in str2
		for ((_, value) in record2)
		{
			operation += value;
		}
		println(" Given (" + str1 + "),(" + str2 + ")");
		// Display number of remove operation to make anagram
		println(" Result : " + operation);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Anagrams = Anagrams();
	/*
	    Case A
	    ------
	    k m e t e o r o i l , r e m f i l l o t e
	    ↓             ↓             ↓     ↓
	    ①            ②             ③    ④
	    ------------------------------------------------
	    Remove 4 characters to make both string is anagram
	*/
	task.makeAnagram("kmeteoroil", "remfillote");
	/*
	    Case B
	    ------
	    C  S I T W I  N  , W  S  O  I  C  N  L  P
	           ↓   ↓             ↓           ↓  ↓
	           ①  ②             ③          ④  ⑤
	    ----------------------------------------------
	    Remove 5 characters to make both string is anagram
	*/
	task.makeAnagram("CSITWIN", "WSOICNLP");
	/*
	    Case C
	    ------
	    Both string is anagram
	*/
	task.makeAnagram("123", "321");
}

Output

 Given (kmeteoroil),(remfillote)
 Result : 4
 Given (CSITWIN),(WSOICNLP)
 Result : 5
 Given (123),(321)
 Result : 0

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