Possible ways to break a string using brackets

Here given code implementation process.

/*
    Java program for
    Possible ways to break a string using brackets 
*/
class Permutation
{
	public String subString(String text, int start, int last)
	{
		String value = "";
		for (int i = start; i <= last; ++i)
		{
			value = value + text.charAt(i);
		}
		return value;
	}
	public void generate(String text, String output, int start, int last)
	{
		if (start == last)
		{
			System.out.println(output);
			return;
		}
		for (int i = start; i < last; ++i)
		{
			generate(text, 
                     output + "(" + subString(text, start, (i)) + ")",
                     i + 1, 
                     last);
		}
	}
	public static void main(String[] args)
	{
		Permutation task = new Permutation();
		String text1 = "123456";
		String text2 = "ABCD";
		// Case 1  
		System.out.println("Given Text : " + text1);
		task.generate(text1, "", 0, text1.length());
		// Case 2  
		System.out.println("\nGiven Text : " + text2);
		task.generate(text2, "", 0, text2.length());
	}
}

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ program for
    Possible ways to break a string using brackets 
*/
class Permutation
{
	public: string subString(string text, int start, int last)
	{
		string value = "";
		for (int i = start; i <= last; ++i)
		{
			value = value  +  text[i];
		}
		return value;
	}
	void generate(string text, string output, int start, int last)
	{
		if (start == last)
		{
			cout << output << endl;
			return;
		}
		for (int i = start; i < last; ++i)
		{
			this->generate(text, 
                 output  +  "("
				 +  this->subString(text, start, (i))  +  ")", 
                           i + 1, last);
		}
	}
};
int main()
{
	Permutation *task = new Permutation();
	string text1 = "123456";
	string text2 = "ABCD";
	// Case 1  
	cout << "Given Text : " << text1 << endl;
	task->generate(text1, "", 0, text1.length());
	// Case 2  
	cout << "\nGiven Text : " << text2 << endl;
	task->generate(text2, "", 0, text2.length());
	return 0;
}

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
// Include namespace system
using System;
/*
    Csharp program for
    Possible ways to break a string using brackets 
*/
public class Permutation
{
	public String subString(String text, int start, int last)
	{
		String value = "";
		for (int i = start; i <= last; ++i)
		{
			value = value + text[i];
		}
		return value;
	}
	public void generate(String text, String output, int start, int last)
	{
		if (start == last)
		{
			Console.WriteLine(output);
			return;
		}
		for (int i = start; i < last; ++i)
		{
			this.generate(text, 
                          output + "(" + this.subString(text, start, (i)) + ")",
                          i + 1, last);
		}
	}
	public static void Main(String[] args)
	{
		Permutation task = new Permutation();
		String text1 = "123456";
		String text2 = "ABCD";
		// Case 1  
		Console.WriteLine("Given Text : " + text1);
		task.generate(text1, "", 0, text1.Length);
		// Case 2  
		Console.WriteLine("\nGiven Text : " + text2);
		task.generate(text2, "", 0, text2.Length);
	}
}

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
package main
import "fmt"
/*
    Go program for
    Possible ways to break a string using brackets 
*/

func subString(text string, start int, last int) string {
	var value string = ""
	for i := start ; i <= last ; i++ {
		value = value + string(text[i])
	}
	return value
}
func generate(text string, 
	output string, 
	start int, 
	last int) {
	if start == last {
		fmt.Println(output)
		return
	}
	for i := start ; i < last ; i++ {
		generate(text, 
			output + "(" + subString(text, start, (i)) + ")", 
			i + 1, last)
	}
}
func main() {
	
	var text1 string = "123456"
	var text2 string = "ABCD"
	// Case 1  
	fmt.Println("Given Text : ", text1)
	generate(text1, "", 0, len(text1))
	// Case 2  
	fmt.Println("\nGiven Text : ", text2)
	generate(text2, "", 0, len(text2))
}

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
<?php
/*
    Php program for
    Possible ways to break a string using brackets 
*/
class Permutation
{
	public	function subString($text, $start, $last)
	{
		$value = "";
		for ($i = $start; $i <= $last; ++$i)
		{
			$value = $value.$text[$i];
		}
		return $value;
	}
	public	function generate($text, $output, $start, $last)
	{
		if ($start == $last)
		{
			echo($output.
				"\n");
			return;
		}
		for ($i = $start; $i < $last; ++$i)
		{
			$this->generate($text, $output.
				"(".$this->subString($text, $start, ($i)).
				")", $i + 1, $last);
		}
	}
}

function main()
{
	$task = new Permutation();
	$text1 = "123456";
	$text2 = "ABCD";
	// Case 1  
	echo("Given Text : ".$text1.
		"\n");
	$task->generate($text1, "", 0, strlen($text1));
	// Case 2  
	echo("\nGiven Text : ".$text2.
		"\n");
	$task->generate($text2, "", 0, strlen($text2));
}
main();

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
/*
    Node JS program for
    Possible ways to break a string using brackets 
*/
class Permutation
{
	subString(text, start, last)
	{
		var value = "";
		for (var i = start; i <= last; ++i)
		{
			value = value + text.charAt(i);
		}
		return value;
	}
	generate(text, output, start, last)
	{
		if (start == last)
		{
			console.log(output);
			return;
		}
		for (var i = start; i < last; ++i)
		{
			this.generate(text, 
                          output + "(" + this.subString(text, start, (i)) + ")",
                          i + 1, last);
		}
	}
}

function main()
{
	var task = new Permutation();
	var text1 = "123456";
	var text2 = "ABCD";
	// Case 1  
	console.log("Given Text : " + text1);
	task.generate(text1, "", 0, text1.length);
	// Case 2  
	console.log("\nGiven Text : " + text2);
	task.generate(text2, "", 0, text2.length);
}
main();

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
#    Python 3 program for
#    Possible ways to break a string using brackets 
class Permutation :
	def subString(self, text, start, last) :
		value = ""
		i = start
		while (i <= last) :
			value = value + str(text[i])
			i += 1
		
		return value
	
	def generate(self, text, output, start, last) :
		if (start == last) :
			print(output)
			return
		
		i = start
		while (i < last) :
			self.generate(text, 
                output + "(" + self.subString(text, start, (i)) + ")", 
                          i + 1, last)
			i += 1
		
	

def main() :
	task = Permutation()
	text1 = "123456"
	text2 = "ABCD"
	#  Case 1  
	print("Given Text : ", text1)
	task.generate(text1, "", 0, len(text1))
	#  Case 2  
	print("\nGiven Text : ", text2)
	task.generate(text2, "", 0, len(text2))

if __name__ == "__main__": main()

Output

Given Text :  123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text :  ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
#    Ruby program for
#    Possible ways to break a string using brackets 
class Permutation 
	def subString(text, start, last) 
		value = ""
		i = start
		while (i <= last) 
			value = value + text[i].to_s
			i += 1
		end

		return value
	end

	def generate(text, output, start, last) 
		if (start == last) 
			print(output, "\n")
			return
		end

		i = start
		while (i < last) 
			self.generate(text, 
                          output + "(" + self.subString(text, start, (i)) + ")",
                          i + 1, last)
			i += 1
		end

	end

end

def main() 
	task = Permutation.new()
	text1 = "123456"
	text2 = "ABCD"
	#  Case 1  
	print("Given Text : ", text1, "\n")
	task.generate(text1, "", 0, text1.length)
	#  Case 2  
	print("\nGiven Text : ", text2, "\n")
	task.generate(text2, "", 0, text2.length)
end

main()

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
import scala.collection.mutable._;
/*
    Scala program for
    Possible ways to break a string using brackets 
*/
class Permutation()
{
	def subString(text: String, 
                  start: Int, 
                  last: Int): String = {
		var value: String = "";
		var i: Int = start;
		while (i <= last)
		{
			value = value + text.charAt(i).toString();
			i += 1;
		}
		return value;
	}
	def generate(
      text: String, 
          output: String, 
             start: Int, 
                 last: Int): Unit = {
		if (start == last)
		{
			println(output);
			return;
		}
		var i: Int = start;
		while (i < last)
		{
			generate(text, 
                     output + "(" + subString(text, start, (i)) + ")",
                     i + 1, last);
			i += 1;
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Permutation = new Permutation();
		var text1: String = "123456";
		var text2: String = "ABCD";
		// Case 1  
		println("Given Text : " + text1);
		task.generate(text1, "", 0, text1.length());
		// Case 2  
		println("\nGiven Text : " + text2);
		task.generate(text2, "", 0, text2.length());
	}
}

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
import Foundation;
/*
    Swift 4 program for
    Possible ways to break a string using brackets 
*/
class Permutation
{
	func subString(_ text: [Character], 
                    _ start: Int, 
                    _ last: Int) -> String
	{
		var value: String = "";
		var i: Int = start;
		while (i <= last)
		{
			value = value + String(text[i]);
			i += 1;
		}
		return value;
	}
	func generate(_ text: String, 
                  _ output: String, 
                  _ start: Int, 
                  _ last: Int)
	{
		if (start == last)
		{
			print(output);
			return;
		}
		var i: Int = start;
      	let arr = Array(text);
		while (i < last)
		{
			self.generate(text, 
                         output + "(" + self.subString(arr, start, (i)) + ")", 
                          i + 1, last);
			i += 1;
		}
	}
}
func main()
{
	let task: Permutation = Permutation();
	let text1: String = "123456";
	let text2: String = "ABCD";
	// Case 1  
	print("Given Text : ", text1);
	task.generate(text1, "", 0, text1.count);
	// Case 2  
	print("\nGiven Text : ", text2);
	task.generate(text2, "", 0, text2.count);
}
main();

Output

Given Text :  123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text :  ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)
/*
    Kotlin program for
    Possible ways to break a string using brackets 
*/
class Permutation
{
	fun subString(text: String, 
                   start: Int, 
                   last: Int): String
	{
		var value: String = "";
		var i: Int = start;
		while (i <= last)
		{
			value = value + text.get(i).toString();
			i += 1;
		}
		return value;
	}
	fun generate(text: String, 
                 output: String, 
                 start: Int, 
                 last: Int): Unit
	{
		if (start == last)
		{
			println(output);
			return;
		}
		var i: Int = start;
		while (i < last)
		{
			this.generate(text, 
                          output + "(" + this.subString(text, start, (i)) + ")",
                          i + 1, last);
			i += 1;
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Permutation = Permutation();
	val text1: String = "123456";
	val text2: String = "ABCD";
	// Case 1  
	println("Given Text : " + text1);
	task.generate(text1, "", 0, text1.length);
	// Case 2  
	println("\nGiven Text : " + text2);
	task.generate(text2, "", 0, text2.length);
}

Output

Given Text : 123456
(1)(2)(3)(4)(5)(6)
(1)(2)(3)(4)(56)
(1)(2)(3)(45)(6)
(1)(2)(3)(456)
(1)(2)(34)(5)(6)
(1)(2)(34)(56)
(1)(2)(345)(6)
(1)(2)(3456)
(1)(23)(4)(5)(6)
(1)(23)(4)(56)
(1)(23)(45)(6)
(1)(23)(456)
(1)(234)(5)(6)
(1)(234)(56)
(1)(2345)(6)
(1)(23456)
(12)(3)(4)(5)(6)
(12)(3)(4)(56)
(12)(3)(45)(6)
(12)(3)(456)
(12)(34)(5)(6)
(12)(34)(56)
(12)(345)(6)
(12)(3456)
(123)(4)(5)(6)
(123)(4)(56)
(123)(45)(6)
(123)(456)
(1234)(5)(6)
(1234)(56)
(12345)(6)
(123456)

Given Text : ABCD
(A)(B)(C)(D)
(A)(B)(CD)
(A)(BC)(D)
(A)(BCD)
(AB)(C)(D)
(AB)(CD)
(ABC)(D)
(ABCD)


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