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Transform one string to another using minimum number of operation

Here given code implementation process.

/*
    C program for
    Transform one string to another using minimum number of operation
*/
#include <stdio.h>
#include <string.h>

void countMinOperation(char *str1, char *str2)
{
	int a = strlen(str1);
	int b = strlen(str2);
	int result = 0;
	if (a != b)
	{
		// Not possible to transform two different length string
		result = -1;
	}
	else
	{
		int frequency[256];
		// Set initial frequency in possible characters
		for (int i = 0; i < 256; ++i)
		{
			frequency[i] = 0;
		}
		// Count character frequency of  str1
		for (int i = 0; i < a; ++i)
		{
			frequency[str1[i]]++;
		}
		// Reduce character frequency of str2
		for (int i = 0; i < b; ++i)
		{
			frequency[str2[i]]--;
		}
		// Check the different characters 
		for (int i = 0; i < 256 && result != -1; ++i)
		{
			if (frequency[i] != 0)
			{
				result = -1;
			}
		}
		int i = a - 1;
		int j = i;
		// When result is not -1 and value of [i] is 
        // greater than or equal to zero.
		// Count resultant operation
		while (i >= 0 && result != -1)
		{
			while (i >= 0 && str1[i] != str2[j])
			{
				// Need to changes
				result++;
				i--;
			}
			if (i >= 0)
			{
				j--;
				i--;
			}
		}
	}
	printf("\n %s  %s", str1, str2);
	if (result == -1)
	{
		printf("\n No solution \n");
	}
	else
	{
		printf("\n Number of operations required is : %d", result);
	}
}
int main(int argc, char const *argv[])
{
	// Test 
	/*
	    care  race
	    -----------
	    care <- Get a and insert at beginning
	    
	    acre  (➀ operation)

	    acre <- Get r and insert at beginning
	    
	    race  (➁ operation)
	    --------------------
	    care is transform into race

	    Result = 2

	*/
	countMinOperation("care", "race");
	/*
	    absent  banets
	    --------------
	    absent <- Get t and insert at beginning
	    
	    tabsen  (➀ operation)

	    tabsen <- Get e and insert at beginning
	    
	    etabsn  (➁ operation)

	    etabsn <- Get n and insert at beginning
	    
	    netabs  (➂ operation)

	    netabs <- Get a and insert at beginning
	    
	    anetbs  (➃ operation)

	    anetbs <- Get b and insert at beginning
	    
	    banets  (➄ operation)

	    --------------------
	    absent is transform into a banets

	    Result = 5

	*/
	countMinOperation("absent", "banets");
	// Similar to above approach
	countMinOperation("LISTEN", "SILENT");
	countMinOperation("abc", "abc");
	countMinOperation("abc", "axc");
	return 0;
}

Output

 care  race
 Number of operations required is : 2
 absent  banets
 Number of operations required is : 5
 LISTEN  SILENT
 Number of operations required is : 5
 abc  abc
 Number of operations required is : 0
 abc  axc
 No solution
// Include header file
#include <iostream>
#include <string>

using namespace std;
// C++ program for
//  Transform one string to another using minimum number of operation
class Transformation
{
	public: void countMinOperation(string str1, string str2)
	{
		int a = str1.length();
		int b = str2.length();
		int result = 0;
		if (a != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			int frequency[256];
			// Set initial frequency in possible characters
			for (int i = 0; i < 256; ++i)
			{
				frequency[i] = 0;
			}
			// Count character frequency of  str1
			for (int i = 0; i < a; ++i)
			{
				frequency[str1[i]]++;
			}
			// Reduce character frequency of str2
			for (int i = 0; i < b; ++i)
			{
				frequency[str2[i]]--;
			}
			// Check the different characters 
			for (int i = 0; i < 256 && result != -1; ++i)
			{
				if (frequency[i] != 0)
				{
					result = -1;
				}
			}
			int i = a - 1;
			int j = i;
			// When result is not -1 and value of [i] is 
			// greater than or equal to zero.
			// Count resultant operation
			while (i >= 0 && result != -1)
			{
				while (i >= 0 && str1[i] != str2[j])
				{
					// Need to changes
					result++;
					i--;
				}
				if (i >= 0)
				{
					j--;
					i--;
				}
			}
		}
		cout << "\n " << str1 << " " << str2;
		if (result == -1)
		{
			cout << "\n No solution \n";
		}
		else
		{
			cout << "\n Number of operations required is : " << result;
		}
	}
};
int main()
{
	Transformation *task = new Transformation();
	// Test 
	/*
			    care  race
			    -----------
			    care <- Get a and insert at beginning
			    
			    acre  (➀ operation)

			    acre <- Get r and insert at beginning
			    
			    race  (➁ operation)
			    --------------------
			    care is transform into race

			    Result = 2

			*/
	task->countMinOperation("care", "race");
	/*
			    absent  banets
			    --------------
			    absent <- Get t and insert at beginning
			    
			    tabsen  (➀ operation)

			    tabsen <- Get e and insert at beginning
			    
			    etabsn  (➁ operation)

			    etabsn <- Get n and insert at beginning
			    
			    netabs  (➂ operation)

			    netabs <- Get a and insert at beginning
			    
			    anetbs  (➃ operation)

			    anetbs <- Get b and insert at beginning
			    
			    banets  (➄ operation)

			    --------------------
			    absent is transform into a banets

			    Result = 5

			*/
	task->countMinOperation("absent", "banets");
	// Similar to above approach
	task->countMinOperation("LISTEN", "SILENT");
	task->countMinOperation("abc", "abc");
	task->countMinOperation("abc", "axc");
	return 0;
}

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
// Java program for
//  Transform one string to another using minimum number of operation
public class Transformation
{
	public void countMinOperation(String str1, String str2)
	{
		int a = str1.length();
		int b = str2.length();
		int result = 0;
		if (a != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			int[] frequency = new int[256];
			// Set initial frequency in possible characters
			for (int i = 0; i < 256; ++i)
			{
				frequency[i] = 0;
			}
			// Count character frequency of  str1
			for (int i = 0; i < a; ++i)
			{
				frequency[str1.charAt(i)]++;
			}
			// Reduce character frequency of str2
			for (int i = 0; i < b; ++i)
			{
				frequency[str2.charAt(i)]--;
			}
			// Check the different characters 
			for (int i = 0; i < 256 && result != -1; ++i)
			{
				if (frequency[i] != 0)
				{
					result = -1;
				}
			}
			int k = a - 1;
			int j = k;
			// When result is not -1 and value of [i] is 
			// greater than or equal to zero.
			// Count resultant operation
			while (k >= 0 && result != -1)
			{
				while (k >= 0 && str1.charAt(k) != str2.charAt(j))
				{
					// Need to changes
					result++;
					k--;
				}
				if (k >= 0)
				{
					j--;
					k--;
				}
			}
		}
		System.out.print("\n " + str1 + " " + str2);
		if (result == -1)
		{
			System.out.print("\n No solution \n");
		}
		else
		{
			System.out.print("\n Number of operations required is : " + result);
		}
	}
	public static void main(String[] args)
	{
		Transformation task = new Transformation();
		// Test 
		/*
		    care  race
		    -----------
		    care <- Get a and insert at beginning
		    
		    acre  (➀ operation)

		    acre <- Get r and insert at beginning
		    
		    race  (➁ operation)
		    --------------------
		    care is transform into race

		    Result = 2

		*/
		task.countMinOperation("care", "race");
		/*
		    absent  banets
		    --------------
		    absent <- Get t and insert at beginning
		    
		    tabsen  (➀ operation)

		    tabsen <- Get e and insert at beginning
		    
		    etabsn  (➁ operation)

		    etabsn <- Get n and insert at beginning
		    
		    netabs  (➂ operation)

		    netabs <- Get a and insert at beginning
		    
		    anetbs  (➃ operation)

		    anetbs <- Get b and insert at beginning
		    
		    banets  (➄ operation)

		    --------------------
		    absent is transform into a banets

		    Result = 5

		*/
		task.countMinOperation("absent", "banets");
		// Similar to above approach
		task.countMinOperation("LISTEN", "SILENT");
		task.countMinOperation("abc", "abc");
		task.countMinOperation("abc", "axc");
	}
}

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
// Include namespace system
using System;
// Csharp program for
//  Transform one string to another using minimum number of operation
public class Transformation
{
	public void countMinOperation(String str1, String str2)
	{
		int a = str1.Length;
		int b = str2.Length;
		int result = 0;
		if (a != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			int[] frequency = new int[256];
			// Set initial frequency in possible characters
			for (int i = 0; i < 256; ++i)
			{
				frequency[i] = 0;
			}
			// Count character frequency of  str1
			for (int i = 0; i < a; ++i)
			{
				frequency[str1[i]]++;
			}
			// Reduce character frequency of str2
			for (int i = 0; i < b; ++i)
			{
				frequency[str2[i]]--;
			}
			// Check the different characters
			for (int i = 0; i < 256 && result != -1; ++i)
			{
				if (frequency[i] != 0)
				{
					result = -1;
				}
			}
			int k = a - 1;
			int j = k;
			// When result is not -1 and value of [i] is
			// greater than or equal to zero.
			// Count resultant operation
			while (k >= 0 && result != -1)
			{
				while (k >= 0 && str1[k] != str2[j])
				{
					// Need to changes
					result++;
					k--;
				}
				if (k >= 0)
				{
					j--;
					k--;
				}
			}
		}
		Console.Write("\n " + str1 + " " + str2);
		if (result == -1)
		{
			Console.Write("\n No solution \n");
		}
		else
		{
			Console.Write("\n Number of operations required is : " + result);
		}
	}
	public static void Main(String[] args)
	{
		Transformation task = new Transformation();
		// Test
		/*
		    care  race
		    -----------
		    care <- Get a and insert at beginning
		    
		    acre  (➀ operation)
		    acre <- Get r and insert at beginning
		    
		    race  (➁ operation)
		    --------------------
		    care is transform into race
		    Result = 2
		*/
		task.countMinOperation("care", "race");
		/*
		    absent  banets
		    --------------
		    absent <- Get t and insert at beginning
		    
		    tabsen  (➀ operation)
		    tabsen <- Get e and insert at beginning
		    
		    etabsn  (➁ operation)
		    etabsn <- Get n and insert at beginning
		    
		    netabs  (➂ operation)
		    netabs <- Get a and insert at beginning
		    
		    anetbs  (➃ operation)
		    anetbs <- Get b and insert at beginning
		    
		    banets  (➄ operation)
		    --------------------
		    absent is transform into a banets
		    Result = 5
		*/
		task.countMinOperation("absent", "banets");
		// Similar to above approach
		task.countMinOperation("LISTEN", "SILENT");
		task.countMinOperation("abc", "abc");
		task.countMinOperation("abc", "axc");
	}
}

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
package main

import "fmt"
// Go program for
// Transform one string to another using minimum number of operation

func countMinOperation(str1, str2 string) {
	var a int = len(str1)
	var b int = len(str2)
	var result int = 0
	if a != b {
		// Not possible to transform two different length string
		result = -1
	} else {
		// Set initial frequency in possible characters
		var frequency = make([] int, 256)
		// Count character frequency of  str1
		for i := 0 ; i < a ; i++ {
			frequency[str1[i]]++
		}
		// Reduce character frequency of str2
		for i := 0 ; i < b ; i++ {
			frequency[str2[i]]--
		}
		// Check the different characters
		for i := 0 ; i < 256 && result != -1 ; i++ {
			if frequency[i] != 0 {
				result = -1
			}
		}
		var k int = a - 1
		var j int = k
		// When result is not -1 and value of [i] is
		// greater than or equal to zero.
		// Count resultant operation
		for (k >= 0 && result != -1) {
			for (k >= 0 && str1[k] != str2[j]) {
				// Need to changes
				result++
				k--
			}
			if k >= 0 {
				j--
				k--
			}
		}
	}
	fmt.Print("\n ", str1, " ", str2)
	if result == -1 {
		fmt.Print("\n No solution \n")
	} else {
		fmt.Print("\n Number of operations required is : ", result)
	}
}
func main() {

	// Test
	/*
	    care  race
	    -----------
	    care <- Get a and insert at beginning
	    
	    acre  (➀ operation)
	    acre <- Get r and insert at beginning
	    
	    race  (➁ operation)
	    --------------------
	    care is transform into race
	    Result = 2
	*/
	countMinOperation("care", "race")
	/*
	    absent  banets
	    --------------
	    absent <- Get t and insert at beginning
	    
	    tabsen  (➀ operation)
	    tabsen <- Get e and insert at beginning
	    
	    etabsn  (➁ operation)
	    etabsn <- Get n and insert at beginning
	    
	    netabs  (➂ operation)
	    netabs <- Get a and insert at beginning
	    
	    anetbs  (➃ operation)
	    anetbs <- Get b and insert at beginning
	    
	    banets  (➄ operation)
	    --------------------
	    absent is transform into a banets
	    Result = 5
	*/
	countMinOperation("absent", "banets")
	// Similar to above approach
	countMinOperation("LISTEN", "SILENT")
	countMinOperation("abc", "abc")
	countMinOperation("abc", "axc")
}

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
<?php
// Php program for
// Transform one string to another using minimum number of operation
class Transformation
{
	public	function countMinOperation($str1, $str2)
	{
		$a = strlen($str1);
		$b = strlen($str2);
		$result = 0;
		if ($a != $b)
		{
			// Not possible to transform two different length string
			$result = -1;
		}
		else
		{
			$frequency = array_fill(0, 256, 0);
			// Set initial frequency in possible characters
			for ($i = 0; $i < 256; ++$i)
			{
				$frequency[$i] = 0;
			}
			// Count character frequency of  str1
			for ($i = 0; $i < $a; ++$i)
			{
				$frequency[ord($str1[$i])]++;
			}
			// Reduce character frequency of str2
			for ($i = 0; $i < $b; ++$i)
			{
				$frequency[ord($str2[$i])]--;
			}
			// Check the different characters
			for ($i = 0; $i < 256 && $result != -1; ++$i)
			{
				if ($frequency[$i] != 0)
				{
					$result = -1;
				}
			}
			$k = $a - 1;
			$j = $k;
			// When result is not -1 and value of [i] is
			// greater than or equal to zero.
			// Count resultant operation
			while ($k >= 0 && $result != -1)
			{
				while ($k >= 0 && $str1[$k] != $str2[$j])
				{
					// Need to changes
					$result++;
					$k--;
				}
				if ($k >= 0)
				{
					$j--;
					$k--;
				}
			}
		}
		echo("\n ".$str1." ".$str2);
		if ($result == -1)
		{
			echo("\n No solution \n");
		}
		else
		{
			echo("\n Number of operations required is : ".$result);
		}
	}
}

function main()
{
	$task = new Transformation();
	// Test
	/*
	    care  race
	    -----------
	    care <- Get a and insert at beginning
	    
	    acre  (➀ operation)
	    acre <- Get r and insert at beginning
	    
	    race  (➁ operation)
	    --------------------
	    care is transform into race
	    Result = 2
	*/
	$task->countMinOperation("care", "race");
	/*
	    absent  banets
	    --------------
	    absent <- Get t and insert at beginning
	    
	    tabsen  (➀ operation)
	    tabsen <- Get e and insert at beginning
	    
	    etabsn  (➁ operation)
	    etabsn <- Get n and insert at beginning
	    
	    netabs  (➂ operation)
	    netabs <- Get a and insert at beginning
	    
	    anetbs  (➃ operation)
	    anetbs <- Get b and insert at beginning
	    
	    banets  (➄ operation)
	    --------------------
	    absent is transform into a banets
	    Result = 5
	*/
	$task->countMinOperation("absent", "banets");
	// Similar to above approach
	$task->countMinOperation("LISTEN", "SILENT");
	$task->countMinOperation("abc", "abc");
	$task->countMinOperation("abc", "axc");
}
main();

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
// Node JS program for
//  Transform one string to another using minimum number of operation
class Transformation
{
	countMinOperation(str1, str2)
	{
		var a = str1.length;
		var b = str2.length;
		var result = 0;
		if (a != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			// Set initial frequency in possible characters
			var frequency = Array(256).fill(0);
			// Count character frequency of  str1
			for (var i = 0; i < a; ++i)
			{
				frequency[str1.charCodeAt(i)]++;
			}
			// Reduce character frequency of str2
			for (var i = 0; i < b; ++i)
			{
				frequency[str2.charCodeAt(i)]--;
			}
			// Check the different characters
			for (var i = 0; i < 256 && result != -1; ++i)
			{
				if (frequency[i] != 0)
				{
					result = -1;
				}
			}
			var k = a - 1;
			var j = k;
			// When result is not -1 and value of [i] is
			// greater than or equal to zero.
			// Count resultant operation
			while (k >= 0 && result != -1)
			{
				while (k >= 0 && str1.charAt(k) != str2.charAt(j))
				{
					// Need to changes
					result++;
					k--;
				}
				if (k >= 0)
				{
					j--;
					k--;
				}
			}
		}
		process.stdout.write("\n " + str1 + " " + str2);
		if (result == -1)
		{
			process.stdout.write("\n No solution \n");
		}
		else
		{
			process.stdout.write("\n Number of operations required is : " + result);
		}
	}
}

function main()
{
	var task = new Transformation();
	// Test
	/*
	    care  race
	    -----------
	    care <- Get a and insert at beginning
	    
	    acre  (➀ operation)
	    acre <- Get r and insert at beginning
	    
	    race  (➁ operation)
	    --------------------
	    care is transform into race
	    Result = 2
	*/
	task.countMinOperation("care", "race");
	/*
	    absent  banets
	    --------------
	    absent <- Get t and insert at beginning
	    
	    tabsen  (➀ operation)
	    tabsen <- Get e and insert at beginning
	    
	    etabsn  (➁ operation)
	    etabsn <- Get n and insert at beginning
	    
	    netabs  (➂ operation)
	    netabs <- Get a and insert at beginning
	    
	    anetbs  (➃ operation)
	    anetbs <- Get b and insert at beginning
	    
	    banets  (➄ operation)
	    --------------------
	    absent is transform into a banets
	    Result = 5
	*/
	task.countMinOperation("absent", "banets");
	// Similar to above approach
	task.countMinOperation("LISTEN", "SILENT");
	task.countMinOperation("abc", "abc");
	task.countMinOperation("abc", "axc");
}
main();

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
#  Python 3 program for
#   Transform one string to another using minimum number of operation
class Transformation :
	def countMinOperation(self, str1, str2) :
		a = len(str1)
		b = len(str2)
		result = 0
		if (a != b) :
			#  Not possible to transform two different length string
			result = -1
		else :
			#  Set initial frequency in possible characters
			frequency = [0] * (256)
			i = 0
			#  Count character frequency of  str1
			while (i < a) :
				frequency[ord(str1[i])] += 1
				i += 1
			
			i = 0
			#  Reduce character frequency of str2
			while (i < b) :
				frequency[ord(str2[i])] -= 1
				i += 1
			
			i = 0
			#  Check the different characters
			while (i < 256 and result != -1) :
				if (frequency[i] != 0) :
					result = -1
				
				i += 1
			
			k = a - 1
			j = k
			#  When result is not -1 and value of [i] is
			#  greater than or equal to zero.
			#  Count resultant operation
			while (k >= 0 and result != -1) :
				while (k >= 0 and str1[k] != str2[j]) :
					#  Need to changes
					result += 1
					k -= 1
				
				if (k >= 0) :
					j -= 1
					k -= 1
				
			
		
		print("\n ", str1 ," ", str2, end = "")
		if (result == -1) :
			print("\n No solution ")
		else :
			print("\n Number of operations required is : ", result, end = "")
		
	

def main() :
	task = Transformation()
	#  Test
	#    care  race
	#    -----------
	#    care <- Get a and insert at beginning
	#    acre  (➀ operation)
	#    acre <- Get r and insert at beginning
	#    race  (➁ operation)
	#    --------------------
	#    care is transform into race
	#    Result = 2
	task.countMinOperation("care", "race")
	#    absent  banets
	#    --------------
	#    absent <- Get t and insert at beginning
	#    tabsen  (➀ operation)
	#    tabsen <- Get e and insert at beginning
	#    etabsn  (➁ operation)
	#    etabsn <- Get n and insert at beginning
	#    netabs  (➂ operation)
	#    netabs <- Get a and insert at beginning
	#    anetbs  (➃ operation)
	#    anetbs <- Get b and insert at beginning
	#    banets  (➄ operation)
	#    --------------------
	#    absent is transform into a banets
	#    Result = 5
	task.countMinOperation("absent", "banets")
	#  Similar to above approach
	task.countMinOperation("LISTEN", "SILENT")
	task.countMinOperation("abc", "abc")
	task.countMinOperation("abc", "axc")

if __name__ == "__main__": main()

Output

  care   race
 Number of operations required is :  2
  absent   banets
 Number of operations required is :  5
  LISTEN   SILENT
 Number of operations required is :  5
  abc   abc
 Number of operations required is :  0
  abc   axc
 No solution
#  Ruby program for
#   Transform one string to another using minimum number of operation
class Transformation 
	def countMinOperation(str1, str2) 
		a = str1.length
		b = str2.length
		result = 0
		if (a != b) 
			#  Not possible to transform two different length string
			result = -1
		else
 
			#  Set initial frequency in possible characters
			frequency = Array.new(256) {0}
			i = 0
			#  Count character frequency of  str1
			while (i < a) 
				frequency[str1[i].ord] += 1
				i += 1
			end

			i = 0
			#  Reduce character frequency of str2
			while (i < b) 
				frequency[str2[i].ord] -= 1
				i += 1
			end

			i = 0
			#  Check the different characters
			while (i < 256 && result != -1) 
				if (frequency[i] != 0) 
					result = -1
				end

				i += 1
			end

			k = a - 1
			j = k
			#  When result is not -1 and value of [i] is
			#  greater than or equal to zero.
			#  Count resultant operation
			while (k >= 0 && result != -1) 
				while (k >= 0 && str1[k] != str2[j]) 
					#  Need to changes
					result += 1
					k -= 1
				end

				if (k >= 0) 
					j -= 1
					k -= 1
				end

			end

		end

		print("\n ", str1 ," ", str2)
		if (result == -1) 
			print("\n No solution \n")
		else
 
			print("\n Number of operations required is : ", result)
		end

	end

end

def main() 
	task = Transformation.new()
	#  Test
	#    care  race
	#    -----------
	#    care <- Get a and insert at beginning
	#    acre  (➀ operation)
	#    acre <- Get r and insert at beginning
	#    race  (➁ operation)
	#    --------------------
	#    care is transform into race
	#    Result = 2
	task.countMinOperation("care", "race")
	#    absent  banets
	#    --------------
	#    absent <- Get t and insert at beginning
	#    tabsen  (➀ operation)
	#    tabsen <- Get e and insert at beginning
	#    etabsn  (➁ operation)
	#    etabsn <- Get n and insert at beginning
	#    netabs  (➂ operation)
	#    netabs <- Get a and insert at beginning
	#    anetbs  (➃ operation)
	#    anetbs <- Get b and insert at beginning
	#    banets  (➄ operation)
	#    --------------------
	#    absent is transform into a banets
	#    Result = 5
	task.countMinOperation("absent", "banets")
	#  Similar to above approach
	task.countMinOperation("LISTEN", "SILENT")
	task.countMinOperation("abc", "abc")
	task.countMinOperation("abc", "axc")
end

main()

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution 

// Scala program for
//  Transform one string to another using minimum number of operation
class Transformation()
{
	def countMinOperation(str1: String, str2: String): Unit = {
		var a: Int = str1.length();
		var b: Int = str2.length();
		var result: Int = 0;
		if (a != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			// Set initial frequency in possible characters
			var frequency: Array[Int] = Array.fill[Int](256)(0);
			var i: Int = 0;
			// Count character frequency of  str1
			while (i < a)
			{
				frequency(str1.charAt(i)) += 1;
				i += 1;
			}
			i = 0;
			// Reduce character frequency of str2
			while (i < b)
			{
				frequency(str2.charAt(i)) -= 1;
				i += 1;
			}
			i = 0;
			// Check the different characters
			while (i < 256 && result != -1)
			{
				if (frequency(i) != 0)
				{
					result = -1;
				}
				i += 1;
			}
			var k: Int = a - 1;
			var j: Int = k;
			// When result is not -1 and value of [i] is
			// greater than or equal to zero.
			// Count resultant operation
			while (k >= 0 && result != -1)
			{
				while (k >= 0 && str1.charAt(k) != str2.charAt(j))
				{
					// Need to changes
					result += 1;
					k -= 1;
				}
				if (k >= 0)
				{
					j -= 1;
					k -= 1;
				}
			}
		}
		print("\n " + str1 + " " + str2);
		if (result == -1)
		{
			print("\n No solution \n");
		}
		else
		{
			print("\n Number of operations required is : " + result);
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Transformation = new Transformation();
		// Test
		/*
		    care  race
		    -----------
		    care <- Get a and insert at beginning
		    
		    acre  (➀ operation)
		    acre <- Get r and insert at beginning
		    
		    race  (➁ operation)
		    --------------------
		    care is transform into race
		    Result = 2
		*/
		task.countMinOperation("care", "race");
		/*
		    absent  banets
		    --------------
		    absent <- Get t and insert at beginning
		    
		    tabsen  (➀ operation)
		    tabsen <- Get e and insert at beginning
		    
		    etabsn  (➁ operation)
		    etabsn <- Get n and insert at beginning
		    
		    netabs  (➂ operation)
		    netabs <- Get a and insert at beginning
		    
		    anetbs  (➃ operation)
		    anetbs <- Get b and insert at beginning
		    
		    banets  (➄ operation)
		    --------------------
		    absent is transform into a banets
		    Result = 5
		*/
		task.countMinOperation("absent", "banets");
		// Similar to above approach
		task.countMinOperation("LISTEN", "SILENT");
		task.countMinOperation("abc", "abc");
		task.countMinOperation("abc", "axc");
	}
}

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
import Foundation;
// Swift 4 program for
//  Transform one string to another using minimum number of operation
class Transformation
{
	func countMinOperation(_ text1: String, _ text2: String)
	{
      	let str1 = Array(text1);
      	let str2 = Array(text2);
		let a: Int = str1.count;
		let b: Int = str2.count;
		var result: Int = 0;
		if (a  != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			// Set initial frequency in possible characters
			var frequency: [Int] = Array(repeating: 0, count: 256);
			var i: Int = 0;
			// Count character frequency of  str1
			while (i < a)
			{
				frequency[Int(UnicodeScalar(String(str1[i]))!.value)] += 1;
				i += 1;
			}
			i = 0;
			// Reduce character frequency of str2
			while (i < b)
			{
				frequency[Int(UnicodeScalar(String(str2[i]))!.value)] -= 1;
				i += 1;
			}
			i = 0;
			// Check the different characters
			while (i < 256 && result  != -1)
			{
				if (frequency[i]  != 0)
				{
					result = -1;
				}
				i += 1;
			}
			var k: Int = a - 1;
			var j: Int = k;
			// When result is not -1 and value of [i] is
			// greater than or equal to zero.
			// Count resultant operation
			while (k >= 0 && result  != -1)
			{
				while (k >= 0 && str1[k]  != str2[j])
				{
					// Need to changes
					result += 1;
					k -= 1;
				}
				if (k >= 0)
				{
					j -= 1;
					k -= 1;
				}
			}
		}
		print("\n ", text1 ," ", text2, terminator: "");
		if (result == -1)
		{
			print("\n No solution ");
		}
		else
		{
			print("\n Number of operations required is : ", 
                  result, terminator: "");
		}
	}
}
func main()
{
	let task: Transformation = Transformation();
	// Test
	/*
	    care  race
	    -----------
	    care <- Get a and insert at beginning
	    
	    acre  (➀ operation)
	    acre <- Get r and insert at beginning
	    
	    race  (➁ operation)
	    --------------------
	    care is transform into race
	    Result = 2
	*/
	task.countMinOperation("care", "race");
	/*
	    absent  banets
	    --------------
	    absent <- Get t and insert at beginning
	    
	    tabsen  (➀ operation)
	    tabsen <- Get e and insert at beginning
	    
	    etabsn  (➁ operation)
	    etabsn <- Get n and insert at beginning
	    
	    netabs  (➂ operation)
	    netabs <- Get a and insert at beginning
	    
	    anetbs  (➃ operation)
	    anetbs <- Get b and insert at beginning
	    
	    banets  (➄ operation)
	    --------------------
	    absent is transform into a banets
	    Result = 5
	*/
	task.countMinOperation("absent", "banets");
	// Similar to above approach
	task.countMinOperation("LISTEN", "SILENT");
	task.countMinOperation("abc", "abc");
	task.countMinOperation("abc", "axc");
}
main();

Output

  care   race
 Number of operations required is :  2
  absent   banets
 Number of operations required is :  5
  LISTEN   SILENT
 Number of operations required is :  5
  abc   abc
 Number of operations required is :  0
  abc   axc
 No solution
// Kotlin program for
//  Transform one string to another using minimum number of operation
class Transformation
{
	fun countMinOperation(str1: String, str2: String): Unit
	{
		val a: Int = str1.length;
		val b: Int = str2.length;
		var result: Int = 0;
		if (a != b)
		{
			// Not possible to transform two different length string
			result = -1;
		}
		else
		{
			// Set initial frequency in possible characters
			val frequency: Array < Int > = Array(256)
			{
				0
			};
			var i: Int = 0;
			// Count character frequency of  str1
			while (i < a)
			{
				frequency[str1.get(i).toInt()] += 1;
				i += 1;
			}
			i = 0;
			// Reduce character frequency of str2
			while (i < b)
			{
				frequency[str2.get(i).toInt()] -= 1;
				i += 1;
			}
			i = 0;
			// Check the different characters
			while (i < 256 && result != -1)
			{
				if (frequency[i] != 0)
				{
					result = -1;
				}
				i += 1;
			}
			var k: Int = a - 1;
			var j: Int = k;
			// When result is not -1 and value of [i] is
			// greater than or equal to zero.
			// Count resultant operation
			while (k >= 0 && result != -1)
			{
				while (k >= 0 && str1.get(k) != str2.get(j))
				{
					// Need to changes
					result += 1;
					k -= 1;
				}
				if (k >= 0)
				{
					j -= 1;
					k -= 1;
				}
			}
		}
		print("\n " + str1 + " " + str2);
		if (result == -1)
		{
			print("\n No solution \n");
		}
		else
		{
			print("\n Number of operations required is : " + result);
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Transformation = Transformation();
	// Test
	/*
	    care  race
	    -----------
	    care <- Get a and insert at beginning
	    
	    acre  (➀ operation)
	    acre <- Get r and insert at beginning
	    
	    race  (➁ operation)
	    --------------------
	    care is transform into race
	    Result = 2
	*/
	task.countMinOperation("care", "race");
	/*
	    absent  banets
	    --------------
	    absent <- Get t and insert at beginning
	    
	    tabsen  (➀ operation)
	    tabsen <- Get e and insert at beginning
	    
	    etabsn  (➁ operation)
	    etabsn <- Get n and insert at beginning
	    
	    netabs  (➂ operation)
	    netabs <- Get a and insert at beginning
	    
	    anetbs  (➃ operation)
	    anetbs <- Get b and insert at beginning
	    
	    banets  (➄ operation)
	    --------------------
	    absent is transform into a banets
	    Result = 5
	*/
	task.countMinOperation("absent", "banets");
	// Similar to above approach
	task.countMinOperation("LISTEN", "SILENT");
	task.countMinOperation("abc", "abc");
	task.countMinOperation("abc", "axc");
}

Output

 care race
 Number of operations required is : 2
 absent banets
 Number of operations required is : 5
 LISTEN SILENT
 Number of operations required is : 5
 abc abc
 Number of operations required is : 0
 abc axc
 No solution
Last updated on by Kalkicode




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