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Minimum number of perpend character needed to Make a string palindrome

Here given code implementation process.

/*
  Java program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
public class Palindrome
{
	// Check that prefix of string is part of palindrome or not
	public boolean isPrefixPalindrome(String text, int n)
	{
		int i = 0;
		int j = n;
		while (i < j)
		{
			if (text.charAt(i) != text.charAt(j))
			{
				// When not a palindrome
				return false;
			}
			i++;
			j--;
		}
		// When yes
		return true;
	}
	public void makePalindromeByPerpend(String text)
	{
		if (text.length() == 0)
		{
			return;
		}
		int count = -1;
		// Get the length of text
		int n = text.length();
		// Executing this loop through by size of text length
		for (int i = n - 1; i >= 0 && count == -1; --i)
		{
			if (text.charAt(0) == text.charAt(i) && 
                isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
		}
		// Display given text
		System.out.println(" Given Text : " + text);
		System.out.println(" Need to Perpend " + count + " Characters");
	}
	public static void main(String[] args)
	{
		Palindrome task = new Palindrome();
		// Test Inputs
		// Case A
		// Text : ipid
		//  d ipi d  [Add d at beginning] 
		// dipid => palindrome
		// Result :1
		task.makePalindromeByPerpend("ipid");
		// Case B
		// Text : 123321892
		// 298 123321 892 [Add 298 at beginning]
		// 298123321892 => palindrome
		// Result : 3
		task.makePalindromeByPerpend("123321892");
		// Case C
		// Text : 121
		// 121 it's already palindrome
		// Result : 0
		task.makePalindromeByPerpend("121");
		// Case D
		// Text : 12356
		// 6532 1 2356 
		// Result : 4
		task.makePalindromeByPerpend("12356");
	}
}

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
  C++ program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
class Palindrome
{
	public:
		// Check that prefix of string is part of palindrome or not
		bool isPrefixPalindrome(string text, int n)
		{
			int i = 0;
			int j = n;
			while (i < j)
			{
				if (text[i] != text[j])
				{
					// When not a palindrome
					return false;
				}
				i++;
				j--;
			}
			// When yes
			return true;
		}
	void makePalindromeByPerpend(string text)
	{
		if (text.length() == 0)
		{
			return;
		}
		int count = -1;
		// Get the length of text
		int n = text.length();
		// Executing this loop through by size of text length
		for (int i = n - 1; i >= 0 && count == -1; --i)
		{
			if (text[0] == text[i] && 
                this->isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
		}
		// Display given text
		cout << " Given Text : " 
            << text << endl;
		cout << " Need to Perpend " 
      		 << count 
             << " Characters" << endl;
	}
};
int main()
{
	Palindrome *task = new Palindrome();
	// Test Inputs
	// Case A
	// Text : ipid
	//  d ipi d  [Add d at beginning] 
	// dipid => palindrome
	// Result :1
	task->makePalindromeByPerpend("ipid");
	// Case B
	// Text : 123321892
	// 298 123321 892 [Add 298 at beginning]
	// 298123321892 => palindrome
	// Result : 3
	task->makePalindromeByPerpend("123321892");
	// Case C
	// Text : 121
	// 121 it's already palindrome
	// Result : 0
	task->makePalindromeByPerpend("121");
	// Case D
	// Text : 12356
	// 6532 1 2356 
	// Result : 4
	task->makePalindromeByPerpend("12356");
	return 0;
}

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
// Include namespace system
using System;
/*
  Csharp program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
public class Palindrome
{
	// Check that prefix of string is part of palindrome or not
	public Boolean isPrefixPalindrome(String text, int n)
	{
		int i = 0;
		int j = n;
		while (i < j)
		{
			if (text[i] != text[j])
			{
				// When not a palindrome
				return false;
			}
			i++;
			j--;
		}
		// When yes
		return true;
	}
	public void makePalindromeByPerpend(String text)
	{
		if (text.Length == 0)
		{
			return;
		}
		int count = -1;
		// Get the length of text
		int n = text.Length;
		// Executing this loop through by size of text length
		for (int i = n - 1; i >= 0 && count == -1; --i)
		{
			if (text[0] == text[i] && 
                this.isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
		}
		// Display given text
		Console.WriteLine(" Given Text : " + text);
		Console.WriteLine(" Need to Perpend " + count + " Characters");
	}
	public static void Main(String[] args)
	{
		Palindrome task = new Palindrome();
		// Test Inputs
		// Case A
		// Text : ipid
		//  d ipi d  [Add d at beginning] 
		// dipid => palindrome
		// Result :1
		task.makePalindromeByPerpend("ipid");
		// Case B
		// Text : 123321892
		// 298 123321 892 [Add 298 at beginning]
		// 298123321892 => palindrome
		// Result : 3
		task.makePalindromeByPerpend("123321892");
		// Case C
		// Text : 121
		// 121 it's already palindrome
		// Result : 0
		task.makePalindromeByPerpend("121");
		// Case D
		// Text : 12356
		// 6532 1 2356 
		// Result : 4
		task.makePalindromeByPerpend("12356");
	}
}

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
package main
import "fmt"
/*
  Go program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/

// Check that prefix of string is part of palindrome or not
func isPrefixPalindrome(text string, n int) bool {
	var i int = 0
	var j int = n
	for (i < j) {
		if text[i] != text[j] {
			// When not a palindrome
			return false
		}
		i++
		j--
	}
	// When yes
	return true
}
func makePalindromeByPerpend(text string) {
	if len(text) == 0 {
		return
	}
	var count int = -1
	// Get the length of text
	var n int = len(text)
	// Executing this loop through by size of text length
	for i := n - 1 ; i >= 0 && count == -1 ; i-- {
		if text[0] == text[i] && isPrefixPalindrome(text, i) {
			// When Prefix 0..i is palindrome
			count = (n - 1) - i
		}
	}
	// Display given text
	fmt.Println(" Given Text : ", text)
	fmt.Println(" Need to Perpend ", count, " Characters")
}
func main() {
	// Test Inputs
	// Case A
	// Text : ipid
	//  d ipi d  [Add d at beginning] 
	// dipid => palindrome
	// Result :1
	makePalindromeByPerpend("ipid")
	// Case B
	// Text : 123321892
	// 298 123321 892 [Add 298 at beginning]
	// 298123321892 => palindrome
	// Result : 3
	makePalindromeByPerpend("123321892")
	// Case C
	// Text : 121
	// 121 it's already palindrome
	// Result : 0
	makePalindromeByPerpend("121")
	// Case D
	// Text : 12356
	// 6532 1 2356 
	// Result : 4
	makePalindromeByPerpend("12356")
}

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
<?php
/*
  Php program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
class Palindrome
{
	// Check that prefix of string is part of palindrome or not
	public	function isPrefixPalindrome($text, $n)
	{
		$i = 0;
		$j = $n;
		while ($i < $j)
		{
			if ($text[$i] != $text[$j])
			{
				// When not a palindrome
				return false;
			}
			$i++;
			$j--;
		}
		// When yes
		return true;
	}
	public	function makePalindromeByPerpend($text)
	{
		if (strlen($text) == 0)
		{
			return;
		}
		$count = -1;
		// Get the length of text
		$n = strlen($text);
		// Executing this loop through by size of text length
		for ($i = $n - 1; $i >= 0 && $count == -1; --$i)
		{
			if ($text[0] == $text[$i] && 
                $this->isPrefixPalindrome($text, $i))
			{
				// When Prefix 0..i is palindrome
				$count = ($n - 1) - $i;
			}
		}
		// Display given text
		echo(" Given Text : ".$text.
			"\n");
		echo(" Need to Perpend ".$count.
			" Characters".
			"\n");
	}
}

function main()
{
	$task = new Palindrome();
	// Test Inputs
	// Case A
	// Text : ipid
	//  d ipi d  [Add d at beginning] 
	// dipid => palindrome
	// Result :1
	$task->makePalindromeByPerpend("ipid");
	// Case B
	// Text : 123321892
	// 298 123321 892 [Add 298 at beginning]
	// 298123321892 => palindrome
	// Result : 3
	$task->makePalindromeByPerpend("123321892");
	// Case C
	// Text : 121
	// 121 it's already palindrome
	// Result : 0
	$task->makePalindromeByPerpend("121");
	// Case D
	// Text : 12356
	// 6532 1 2356 
	// Result : 4
	$task->makePalindromeByPerpend("12356");
}
main();

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
/*
  Node JS program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
class Palindrome
{
	// Check that prefix of string is part of palindrome or not
	isPrefixPalindrome(text, n)
	{
		var i = 0;
		var j = n;
		while (i < j)
		{
			if (text.charAt(i) != text.charAt(j))
			{
				// When not a palindrome
				return false;
			}
			i++;
			j--;
		}
		// When yes
		return true;
	}
	makePalindromeByPerpend(text)
	{
		if (text.length == 0)
		{
			return;
		}
		var count = -1;
		// Get the length of text
		var n = text.length;
		// Executing this loop through by size of text length
		for (var i = n - 1; i >= 0 && count == -1; --i)
		{
			if (text.charAt(0) == text.charAt(i) && 
                this.isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
		}
		// Display given text
		console.log(" Given Text : " + text);
		console.log(" Need to Perpend " + count + " Characters");
	}
}

function main()
{
	var task = new Palindrome();
	// Test Inputs
	// Case A
	// Text : ipid
	//  d ipi d  [Add d at beginning] 
	// dipid => palindrome
	// Result :1
	task.makePalindromeByPerpend("ipid");
	// Case B
	// Text : 123321892
	// 298 123321 892 [Add 298 at beginning]
	// 298123321892 => palindrome
	// Result : 3
	task.makePalindromeByPerpend("123321892");
	// Case C
	// Text : 121
	// 121 it's already palindrome
	// Result : 0
	task.makePalindromeByPerpend("121");
	// Case D
	// Text : 12356
	// 6532 1 2356 
	// Result : 4
	task.makePalindromeByPerpend("12356");
}
main();

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
#  Python 3 program for
#  Minimum number of perpend character needed to 
#  Make a string palindrome
class Palindrome :
	#  Check that prefix of string is part of palindrome or not
	def isPrefixPalindrome(self, text, n) :
		i = 0
		j = n
		while (i < j) :
			if (text[i] != text[j]) :
				#  When not a palindrome
				return False
			
			i += 1
			j -= 1
		
		#  When yes
		return True
	
	def makePalindromeByPerpend(self, text) :
		if (len(text) == 0) :
			return
		
		count = -1
		#  Get the length of text
		n = len(text)
		i = n - 1
		#  Executing this loop through by size of text length
		while (i >= 0 and count == -1) :
			if (text[0] == text[i] and 
                self.isPrefixPalindrome(text, i)) :
				#  When Prefix 0..i is palindrome
				count = (n - 1) - i
			
			i -= 1
		
		#  Display given text
		print(" Given Text : ", text)
		print(" Need to Perpend ", count ," Characters")
	

def main() :
	task = Palindrome()
	#  Test Inputs
	#  Case A
	#  Text : ipid
	#   d ipi d  [Add d at beginning] 
	#  dipid => palindrome
	#  Result :1
	task.makePalindromeByPerpend("ipid")
	#  Case B
	#  Text : 123321892
	#  298 123321 892 [Add 298 at beginning]
	#  298123321892 => palindrome
	#  Result : 3
	task.makePalindromeByPerpend("123321892")
	#  Case C
	#  Text : 121
	#  121 it's already palindrome
	#  Result : 0
	task.makePalindromeByPerpend("121")
	#  Case D
	#  Text : 12356
	#  6532 1 2356 
	#  Result : 4
	task.makePalindromeByPerpend("12356")

if __name__ == "__main__": main()

Output

 Given Text :  ipid
 Need to Perpend  1  Characters
 Given Text :  123321892
 Need to Perpend  3  Characters
 Given Text :  121
 Need to Perpend  0  Characters
 Given Text :  12356
 Need to Perpend  4  Characters
#  Ruby program for
#  Minimum number of perpend character needed to 
#  Make a string palindrome

class Palindrome 
	#  Check that prefix of string is part of palindrome or not
	def isPrefixPalindrome(text, n) 
		i = 0
		j = n
		while (i < j) 
			if (text[i] != text[j]) 
				#  When not a palindrome
				return false
			end

			i += 1
			j -= 1
		end

		#  When yes
		return true
	end

	def makePalindromeByPerpend(text) 
		if (text.length == 0) 
			return
		end

		count = -1
		#  Get the length of text
		n = text.length
		i = n - 1
		#  Executing this loop through by size of text length
		while (i >= 0 && count == -1) 
			if (text[0] == text[i] && 
                self.isPrefixPalindrome(text, i)) 
				#  When Prefix 0..i is palindrome
				count = (n - 1) - i
			end

			i -= 1
		end

		#  Display given text
		print(" Given Text : ", text, "\n")
		print(" Need to Perpend ", count ," Characters", "\n")
	end

end

def main() 
	task = Palindrome.new()
	#  Test Inputs
	#  Case A
	#  Text : ipid
	#   d ipi d  [Add d at beginning] 
	#  dipid => palindrome
	#  Result :1
	task.makePalindromeByPerpend("ipid")
	#  Case B
	#  Text : 123321892
	#  298 123321 892 [Add 298 at beginning]
	#  298123321892 => palindrome
	#  Result : 3
	task.makePalindromeByPerpend("123321892")
	#  Case C
	#  Text : 121
	#  121 it's already palindrome
	#  Result : 0
	task.makePalindromeByPerpend("121")
	#  Case D
	#  Text : 12356
	#  6532 1 2356 
	#  Result : 4
	task.makePalindromeByPerpend("12356")
end

main()

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters

import scala.collection.mutable._;
/*
  Scala program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
class Palindrome()
{
	// Check that prefix of string is part of palindrome or not
	def isPrefixPalindrome(text: String, n: Int): Boolean = {
		var i: Int = 0;
		var j: Int = n;
		while (i < j)
		{
			if (text.charAt(i) != text.charAt(j))
			{
				// When not a palindrome
				return false;
			}
			i += 1;
			j -= 1;
		}
		// When yes
		return true;
	}
	def makePalindromeByPerpend(text: String): Unit = {
		if (text.length() == 0)
		{
			return;
		}
		var count: Int = -1;
		// Get the length of text
		var n: Int = text.length();
		var i: Int = n - 1;
		// Executing this loop through by size of text length
		while (i >= 0 && count == -1)
		{
			if (text.charAt(0) == text.charAt(i) && 
                isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
			i -= 1;
		}
		// Display given text
		println(" Given Text : " + text);
		println(" Need to Perpend " + count + " Characters");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Palindrome = new Palindrome();
		// Test Inputs
		// Case A
		// Text : ipid
		//  d ipi d  [Add d at beginning] 
		// dipid => palindrome
		// Result :1
		task.makePalindromeByPerpend("ipid");
		// Case B
		// Text : 123321892
		// 298 123321 892 [Add 298 at beginning]
		// 298123321892 => palindrome
		// Result : 3
		task.makePalindromeByPerpend("123321892");
		// Case C
		// Text : 121
		// 121 it's already palindrome
		// Result : 0
		task.makePalindromeByPerpend("121");
		// Case D
		// Text : 12356
		// 6532 1 2356 
		// Result : 4
		task.makePalindromeByPerpend("12356");
	}
}

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
import Foundation;
/*
  Swift 4 program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
class Palindrome
{
	// Check that prefix of string is part of palindrome or not
	func isPrefixPalindrome(_ text: [Character], _ n: Int) -> Bool
	{
		var i: Int = 0;
		var j: Int = n;
		while (i < j)
		{
			if (text[i]  != text[j])
			{
				// When not a palindrome
				return false;
			}
			i += 1;
			j -= 1;
		}
		// When yes
		return true;
	}
	func makePalindromeByPerpend(_ data: String)
	{
		if (data.count == 0)
		{
			return;
		}
		var count: Int = -1;
		// Get the length of text
		let n: Int = data.count;
		var i: Int = n - 1;
        let text = Array(data);
		// Executing this loop through by size of text length
		while (i >= 0 && count == -1)
		{
			if (text[0] == text[i] && 
                self.isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
			i -= 1;
		}
		// Display given text
		print(" Given Text : ", data);
		print(" Need to Perpend", count ,"Characters");
	}
}
func main()
{
	let task: Palindrome = Palindrome();
	// Test Inputs
	// Case A
	// Text : ipid
	//  d ipi d  [Add d at beginning] 
	// dipid => palindrome
	// Result :1
	task.makePalindromeByPerpend("ipid");
	// Case B
	// Text : 123321892
	// 298 123321 892 [Add 298 at beginning]
	// 298123321892 => palindrome
	// Result : 3
	task.makePalindromeByPerpend("123321892");
	// Case C
	// Text : 121
	// 121 it's already palindrome
	// Result : 0
	task.makePalindromeByPerpend("121");
	// Case D
	// Text : 12356
	// 6532 1 2356 
	// Result : 4
	task.makePalindromeByPerpend("12356");
}
main();

Output

 Given Text :  ipid
 Need to Perpend 1 Characters
 Given Text :  123321892
 Need to Perpend 3 Characters
 Given Text :  121
 Need to Perpend 0 Characters
 Given Text :  12356
 Need to Perpend 4 Characters
/*
  Kotlin program for
  Minimum number of perpend character needed to 
  Make a string palindrome
*/
class Palindrome
{
	// Check that prefix of string is part of palindrome or not
	fun isPrefixPalindrome(text: String, n: Int): Boolean
	{
		var i: Int = 0;
		var j: Int = n;
		while (i < j)
		{
			if (text.get(i) != text.get(j))
			{
				// When not a palindrome
				return false;
			}
			i += 1;
			j -= 1;
		}
		// When yes
		return true;
	}
	fun makePalindromeByPerpend(text: String): Unit
	{
		if (text.length == 0)
		{
			return;
		}
		var count: Int = -1;
		// Get the length of text
		val n: Int = text.length;
		var i: Int = n - 1;
		// Executing this loop through by size of text length
		while (i >= 0 && count == -1)
		{
			if (text.get(0) == text.get(i) && 
                this.isPrefixPalindrome(text, i))
			{
				// When Prefix 0..i is palindrome
				count = (n - 1) - i;
			}
			i -= 1;
		}
		// Display given text
		println(" Given Text : " + text);
		println(" Need to Perpend " + count + " Characters");
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Palindrome = Palindrome();
	// Test Inputs
	// Case A
	// Text : ipid
	//  d ipi d  [Add d at beginning] 
	// dipid => palindrome
	// Result :1
	task.makePalindromeByPerpend("ipid");
	// Case B
	// Text : 123321892
	// 298 123321 892 [Add 298 at beginning]
	// 298123321892 => palindrome
	// Result : 3
	task.makePalindromeByPerpend("123321892");
	// Case C
	// Text : 121
	// 121 it's already palindrome
	// Result : 0
	task.makePalindromeByPerpend("121");
	// Case D
	// Text : 12356
	// 6532 1 2356 
	// Result : 4
	task.makePalindromeByPerpend("12356");
}

Output

 Given Text : ipid
 Need to Perpend 1 Characters
 Given Text : 123321892
 Need to Perpend 3 Characters
 Given Text : 121
 Need to Perpend 0 Characters
 Given Text : 12356
 Need to Perpend 4 Characters
Last updated on by Kalkicode




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