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Longest palindromic substring recursive

Here given code implementation process.

/*
    C program for
    Longest palindromic substring recursive
*/
#include <stdio.h>
#include <string.h>

int maxValue(int a, int b)
{
	if (a > b)
	{
		return a;
	}
	return b;
}
int maxLength(char *text, int s, int e, int length)
{
	if (s == e)
	{
		return length + 1;
	}
	if (s > e)
	{
		return length;
	}
	if (text[s] == text[e])
	{
		// When boundary element are similar.
		// Find the palindrome length in possible three cases.
		// Inner substring, start+1 to end and start to end-1
		return maxValue(
          maxLength(text, s + 1, e - 1, length + 2), 
          maxValue(maxLength(text, s + 1, e, 0), 
                   maxLength(text, s, e - 1, 0)));
	}
	return maxValue(maxLength(text, s + 1, e, 0), 
                    maxLength(text, s, e - 1, 0));
}
void lengthOfLPS(char *text)
{
	// Get length of text
	int n = strlen(text);
	if (n == 0)
	{
		// When text is empty
		return;
	}
	// Display given text
	printf("\n Given string : %s", text);
	// 1st parameters are passing text 
	// 2nd parameter is start point
	// 3rd parameter is last point
	// 4th is resultant length
	int length = maxLength(text, 0, n - 1, 0);
	// Display length of resultant palindrome
	printf("\n Resultant palindrome length : %d", length);
}
int main(int argc, char
	const *argv[])
{
	// Test
	// iiii => 4
	lengthOfLPS("nitiiiigi");
	// ivfvi => 5
	lengthOfLPS("vvvvivfvim");
	// civic => 5
	lengthOfLPS("xyzacivicxrs");
	// opo => 3
	lengthOfLPS("ttoopo");
	return 0;
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
// Java program for
// Longest palindromic substring recursive
public class LongestPalindrome
{
    public int maxValue(int a, int b)
    {
        if (a > b)
        {
            return a;
        }
        return b;
    }
    public int maxPalindromicLength(String text, int s, int e, int length)
    {
        if (s == e)
        {
            return length + 1;
        }
        if (s > e)
        {
            return length;
        }
        if (text.charAt(s) == text.charAt(e))
        {
            // When boundary element are similar.
            // Find the palindrome length in possible three cases.
            // Inner substring, start+1 to end and start to end-1
            return maxValue(maxPalindromicLength(text, s + 1, e - 1, length + 2), 
                maxValue(maxPalindromicLength(text, s + 1, e, 0), 
                    maxPalindromicLength(text, s, e - 1, 0)));
        }
        return maxValue(maxPalindromicLength(text, s + 1, e, 0), 
                        maxPalindromicLength(text, s, e - 1, 0));
    }

    public void lengthOfLPS(String text)
    {
        // Get length of text
        int n = text.length();
        if (n == 0)
        {
            // When text is empty
            return;
        }
        // Display given text
        System.out.print("\n Given string : " + text );
        // 1st parameters are passing text 
        // 2nd parameter is start point
        // 3rd parameter is last point
        // 4th is resultant length
        int length = maxPalindromicLength(text, 0, n - 1, 0);
        // Display length of resultant palindrome
        System.out.print("\n Resultant palindrome length : " + length );
    }
    public static void main(String[] args)
    {
        LongestPalindrome task = new LongestPalindrome();
        // Test
        // iiii => 4
        task.lengthOfLPS("nitiiiigi");
        // ivfvi => 5
        task.lengthOfLPS("vvvvivfvim");
        // civic => 5
        task.lengthOfLPS("xyzacivicxrs");
        // opo => 3
        task.lengthOfLPS("ttoopo");
    }
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
// Include header file
#include <iostream>
#include <string>

using namespace std;
// C++ program for
// Longest palindromic substring recursive
class LongestPalindrome
{
	public: int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	int maxPalindromicLength(string text, int s, int e, int length)
	{
		if (s == e)
		{
			return length + 1;
		}
		if (s > e)
		{
			return length;
		}
		if (text[s] == text[e])
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return this->maxValue(
              this->maxPalindromicLength(text, s + 1, e - 1, length + 2), 
              this->maxValue(this->maxPalindromicLength(text, s + 1, e, 0), 
                             this->maxPalindromicLength(text, s, e - 1, 0)));
		}
		return this->maxValue(this->maxPalindromicLength(text, s + 1, e, 0),
                              this->maxPalindromicLength(text, s, e - 1, 0));
	}
	void lengthOfLPS(string text)
	{
		// Get length of text
		int n = text.length();
		if (n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		cout << "\n Given string : " << text;
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		int length = this->maxPalindromicLength(text, 0, n - 1, 0);
		// Display length of resultant palindrome
		cout << "\n Resultant palindrome length : " << length;
	}
};
int main()
{
	LongestPalindrome *task = new LongestPalindrome();
	// Test
	// iiii => 4
	task->lengthOfLPS("nitiiiigi");
	// ivfvi => 5
	task->lengthOfLPS("vvvvivfvim");
	// civic => 5
	task->lengthOfLPS("xyzacivicxrs");
	// opo => 3
	task->lengthOfLPS("ttoopo");
	return 0;
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
// Include namespace system
using System;
// Csharp program for
// Longest palindromic substring recursive
public class LongestPalindrome
{
	public int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	public int maxPalindromicLength(String text, int s, int e, int length)
	{
		if (s == e)
		{
			return length + 1;
		}
		if (s > e)
		{
			return length;
		}
		if (text[s] == text[e])
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return this.maxValue(
              this.maxPalindromicLength(text, s + 1, e - 1, length + 2), 
              this.maxValue(this.maxPalindromicLength(text, s + 1, e, 0), 
                            this.maxPalindromicLength(text, s, e - 1, 0)));
		}
		return this.maxValue(this.maxPalindromicLength(text, s + 1, e, 0),
                             this.maxPalindromicLength(text, s, e - 1, 0));
	}
	public void lengthOfLPS(String text)
	{
		// Get length of text
		int n = text.Length;
		if (n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		Console.Write("\n Given string : " + text);
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		int length = this.maxPalindromicLength(text, 0, n - 1, 0);
		// Display length of resultant palindrome
		Console.Write("\n Resultant palindrome length : " + length);
	}
	public static void Main(String[] args)
	{
		LongestPalindrome task = new LongestPalindrome();
		// Test
		// iiii => 4
		task.lengthOfLPS("nitiiiigi");
		// ivfvi => 5
		task.lengthOfLPS("vvvvivfvim");
		// civic => 5
		task.lengthOfLPS("xyzacivicxrs");
		// opo => 3
		task.lengthOfLPS("ttoopo");
	}
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
package main
import "fmt"
// Go program for
// Longest palindromic substring recursive

func maxValue(a, b int) int {
	if a > b {
		return a
	}
	return b
}
func maxPalindromicLength(text string, s int, e int, length int) int {
	if s == e {
		return length + 1
	}
	if s > e {
		return length
	}
	if text[s] == text[e] {
		// When boundary element are similar.
		// Find the palindrome length in possible three cases.
		// Inner substring, start+1 to end and start to end-1
		return maxValue(maxPalindromicLength(text, s + 1, e - 1, length + 2), 
			maxValue(maxPalindromicLength(text, s + 1, e, 0), 
				maxPalindromicLength(text, s, e - 1, 0)))
	}
	return maxValue(maxPalindromicLength(text, s + 1, e, 0), 
		maxPalindromicLength(text, s, e - 1, 0))
}
func lengthOfLPS(text string) {
	// Get length of text
	var n int = len(text)
	if n == 0 {
		// When text is empty
		return
	}
	// Display given text
	fmt.Print("\n Given string : ", text)
	// 1st parameters are passing text 
	// 2nd parameter is start point
	// 3rd parameter is last point
	// 4th is resultant length
	var length int = maxPalindromicLength(text, 0, n - 1, 0)
	// Display length of resultant palindrome
	fmt.Print("\n Resultant palindrome length : ", length)
}
func main() {
	
	// Test
	// iiii => 4
	lengthOfLPS("nitiiiigi")
	// ivfvi => 5
	lengthOfLPS("vvvvivfvim")
	// civic => 5
	lengthOfLPS("xyzacivicxrs")
	// opo => 3
	lengthOfLPS("ttoopo")
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
<?php
// Php program for
// Longest palindromic substring recursive
class LongestPalindrome
{
	public	function maxValue($a, $b)
	{
		if ($a > $b)
		{
			return $a;
		}
		return $b;
	}
	public	function maxPalindromicLength($text, $s, $e, $length)
	{
		if ($s == $e)
		{
			return $length + 1;
		}
		if ($s > $e)
		{
			return $length;
		}
		if ($text[$s] == $text[$e])
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return $this->maxValue(
              $this->maxPalindromicLength($text, $s + 1, $e - 1, $length + 2), 
              $this->maxValue($this->maxPalindromicLength($text, $s + 1, $e, 0),
                              $this->maxPalindromicLength($text, $s, $e - 1, 0)));
		}
		return $this->maxValue($this->maxPalindromicLength($text, $s + 1, $e, 0),
                               $this->maxPalindromicLength($text, $s, $e - 1, 0));
	}
	public	function lengthOfLPS($text)
	{
		// Get length of text
		$n = strlen($text);
		if ($n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		echo("\n Given string : ".$text);
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		$length = $this->maxPalindromicLength($text, 0, $n - 1, 0);
		// Display length of resultant palindrome
		echo("\n Resultant palindrome length : ".$length);
	}
}

function main()
{
	$task = new LongestPalindrome();
	// Test
	// iiii => 4
	$task->lengthOfLPS("nitiiiigi");
	// ivfvi => 5
	$task->lengthOfLPS("vvvvivfvim");
	// civic => 5
	$task->lengthOfLPS("xyzacivicxrs");
	// opo => 3
	$task->lengthOfLPS("ttoopo");
}
main();

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
// Node JS program for
// Longest palindromic substring recursive
class LongestPalindrome
{
	maxValue(a, b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	maxPalindromicLength(text, s, e, length)
	{
		if (s == e)
		{
			return length + 1;
		}
		if (s > e)
		{
			return length;
		}
		if (text.charAt(s) == text.charAt(e))
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return this.maxValue(
              this.maxPalindromicLength(text, s + 1, e - 1, length + 2),
              this.maxValue(this.maxPalindromicLength(text, s + 1, e, 0),
                            this.maxPalindromicLength(text, s, e - 1, 0)));
		}
		return this.maxValue(this.maxPalindromicLength(text, s + 1, e, 0),
                             this.maxPalindromicLength(text, s, e - 1, 0));
	}
	lengthOfLPS(text)
	{
		// Get length of text
		var n = text.length;
		if (n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		process.stdout.write("\n Given string : " + text);
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		var length = this.maxPalindromicLength(text, 0, n - 1, 0);
		// Display length of resultant palindrome
		process.stdout.write("\n Resultant palindrome length : " + length);
	}
}

function main()
{
	var task = new LongestPalindrome();
	// Test
	// iiii => 4
	task.lengthOfLPS("nitiiiigi");
	// ivfvi => 5
	task.lengthOfLPS("vvvvivfvim");
	// civic => 5
	task.lengthOfLPS("xyzacivicxrs");
	// opo => 3
	task.lengthOfLPS("ttoopo");
}
main();

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
#  Python 3 program for
#  Longest palindromic substring recursive
class LongestPalindrome :
	def maxValue(self, a, b) :
		if (a > b) :
			return a
		
		return b
	
	def maxPalindromicLength(self, text, s, e, length) :
		if (s == e) :
			return length + 1
		
		if (s > e) :
			return length
		
		if (text[s] == text[e]) :
			#  When boundary element are similar.
			#  Find the palindrome length in possible three cases.
			#  Inner substring, start+1 to end and start to end-1
			return self.maxValue(
              self.maxPalindromicLength(text, s + 1, e - 1, length + 2),
              self.maxValue(self.maxPalindromicLength(text, s + 1, e, 0),
                            self.maxPalindromicLength(text, s, e - 1, 0)))
		
		return self.maxValue(self.maxPalindromicLength(text, s + 1, e, 0),
                             self.maxPalindromicLength(text, s, e - 1, 0))
	
	def lengthOfLPS(self, text) :
		#  Get length of text
		n = len(text)
		if (n == 0) :
			#  When text is empty
			return
		
		#  Display given text
		print("\n Given string : ", text, end = "")
		#  1st parameters are passing text 
		#  2nd parameter is start point
		#  3rd parameter is last point
		#  4th is resultant length
		length = self.maxPalindromicLength(text, 0, n - 1, 0)
		#  Display length of resultant palindrome
		print("\n Resultant palindrome length : ", length, end = "")
	

def main() :
	task = LongestPalindrome()
	#  Test
	#  iiii => 4
	task.lengthOfLPS("nitiiiigi")
	#  ivfvi => 5
	task.lengthOfLPS("vvvvivfvim")
	#  civic => 5
	task.lengthOfLPS("xyzacivicxrs")
	#  opo => 3
	task.lengthOfLPS("ttoopo")

if __name__ == "__main__": main()

Output

 Given string :  nitiiiigi
 Resultant palindrome length :  4
 Given string :  vvvvivfvim
 Resultant palindrome length :  5
 Given string :  xyzacivicxrs
 Resultant palindrome length :  5
 Given string :  ttoopo
 Resultant palindrome length :  3
#  Ruby program for
#  Longest palindromic substring recursive
class LongestPalindrome 
	def maxValue(a, b) 
		if (a > b) 
			return a
		end

		return b
	end

	def maxPalindromicLength(text, s, e, length) 
		if (s == e) 
			return length + 1
		end

		if (s > e) 
			return length
		end

		if (text[s] == text[e]) 
			#  When boundary element are similar.
			#  Find the palindrome length in possible three cases.
			#  Inner substring, start+1 to end and start to end-1
			return self.maxValue(
              self.maxPalindromicLength(text, s + 1, e - 1, length + 2),
              self.maxValue(self.maxPalindromicLength(text, s + 1, e, 0),
                            self.maxPalindromicLength(text, s, e - 1, 0))
            )
		end

		return self.maxValue(
          self.maxPalindromicLength(text, s + 1, e, 0),
          self.maxPalindromicLength(text, s, e - 1, 0))
	end

	def lengthOfLPS(text) 
		#  Get length of text
		n = text.length
		if (n == 0) 
			#  When text is empty
			return
		end

		#  Display given text
		print("\n Given string : ", text)
		#  1st parameters are passing text 
		#  2nd parameter is start point
		#  3rd parameter is last point
		#  4th is resultant length
		length = self.maxPalindromicLength(text, 0, n - 1, 0)
		#  Display length of resultant palindrome
		print("\n Resultant palindrome length : ", length)
	end

end

def main() 
	task = LongestPalindrome.new()
	#  Test
	#  iiii => 4
	task.lengthOfLPS("nitiiiigi")
	#  ivfvi => 5
	task.lengthOfLPS("vvvvivfvim")
	#  civic => 5
	task.lengthOfLPS("xyzacivicxrs")
	#  opo => 3
	task.lengthOfLPS("ttoopo")
end

main()

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
import scala.collection.mutable._;
// Scala program for
// Longest palindromic substring recursive
class LongestPalindrome()
{
	def maxValue(a: Int, b: Int): Int = {
		if (a > b)
		{
			return a;
		}
		return b;
	}
	def maxPalindromicLength(
      		text: String, 
      		s: Int, 
      		e: Int, 
      		length: Int): Int = {
		if (s == e)
		{
			return length + 1;
		}
		if (s > e)
		{
			return length;
		}
		if (text.charAt(s) == text.charAt(e))
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return maxValue(
              maxPalindromicLength(text, s + 1, e - 1, length + 2),
              maxValue(maxPalindromicLength(text, s + 1, e, 0), 
                       maxPalindromicLength(text, s, e - 1, 0)));
		}
		return maxValue(maxPalindromicLength(text, s + 1, e, 0), maxPalindromicLength(text, s, e - 1, 0));
	}
	def lengthOfLPS(text: String): Unit = {
		// Get length of text
		var n: Int = text.length();
		if (n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		print("\n Given string : " + text);
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		var length: Int = maxPalindromicLength(text, 0, n - 1, 0);
		// Display length of resultant palindrome
		print("\n Resultant palindrome length : " + length);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: LongestPalindrome = new LongestPalindrome();
		// Test
		// iiii => 4
		task.lengthOfLPS("nitiiiigi");
		// ivfvi => 5
		task.lengthOfLPS("vvvvivfvim");
		// civic => 5
		task.lengthOfLPS("xyzacivicxrs");
		// opo => 3
		task.lengthOfLPS("ttoopo");
	}
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
import Foundation;
// Swift 4 program for
// Longest palindromic substring recursive
class LongestPalindrome
{
	func maxValue(_ a: Int, _ b: Int) -> Int
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	func maxPalindromicLength(_ text: [Character], 
      						  _ s: Int, 
                              _ e: Int, 
                              _ length: Int) -> Int
	{
		if (s == e)
		{
			return length + 1;
		}
		if (s > e)
		{
			return length;
		}
		if (text[s] == text[e])
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return self.maxValue(
              self.maxPalindromicLength(text, s + 1, e - 1, length + 2),
              self.maxValue(self.maxPalindromicLength(text, s + 1, e, 0),
                            self.maxPalindromicLength(text, s, e - 1, 0)));
		}
		return self.maxValue(self.maxPalindromicLength(text, s + 1, e, 0),
                             self.maxPalindromicLength(text, s, e - 1, 0));
	}
	func lengthOfLPS(_ text: String)
	{
		// Get length of text
		let n: Int = text.count;
		if (n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		print("\n Given string : ", text, terminator: "");
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		let length: Int = self.maxPalindromicLength(Array(text), 0, n - 1, 0);
		// Display length of resultant palindrome
		print("\n Resultant palindrome length : ", length, terminator: "");
	}
}
func main()
{
	let task: LongestPalindrome = LongestPalindrome();
	// Test
	// iiii => 4
	task.lengthOfLPS("nitiiiigi");
	// ivfvi => 5
	task.lengthOfLPS("vvvvivfvim");
	// civic => 5
	task.lengthOfLPS("xyzacivicxrs");
	// opo => 3
	task.lengthOfLPS("ttoopo");
}
main();

Output

 Given string :  nitiiiigi
 Resultant palindrome length :  4
 Given string :  vvvvivfvim
 Resultant palindrome length :  5
 Given string :  xyzacivicxrs
 Resultant palindrome length :  5
 Given string :  ttoopo
 Resultant palindrome length :  3
// Kotlin program for
// Longest palindromic substring recursive
class LongestPalindrome
{
	fun maxValue(a: Int, b: Int): Int
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	fun maxPalindromicLength(text: String, s: Int, e: Int, length: Int): Int
	{
		if (s == e)
		{
			return length + 1;
		}
		if (s > e)
		{
			return length;
		}
		if (text.get(s) == text.get(e))
		{
			// When boundary element are similar.
			// Find the palindrome length in possible three cases.
			// Inner substring, start+1 to end and start to end-1
			return this.maxValue(
              this.maxPalindromicLength(text, s + 1, e - 1, length + 2),
              this.maxValue(this.maxPalindromicLength(text, s + 1, e, 0),
                            this.maxPalindromicLength(text, s, e - 1, 0)));
		}
		return this.maxValue(this.maxPalindromicLength(text, s + 1, e, 0),
                             this.maxPalindromicLength(text, s, e - 1, 0));
	}
	fun lengthOfLPS(text: String): Unit
	{
		// Get length of text
		val n: Int = text.length;
		if (n == 0)
		{
			// When text is empty
			return;
		}
		// Display given text
		print("\n Given string : " + text);
		// 1st parameters are passing text 
		// 2nd parameter is start point
		// 3rd parameter is last point
		// 4th is resultant length
		val length: Int = this.maxPalindromicLength(text, 0, n - 1, 0);
		// Display length of resultant palindrome
		print("\n Resultant palindrome length : " + length);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: LongestPalindrome = LongestPalindrome();
	// Test
	// iiii => 4
	task.lengthOfLPS("nitiiiigi");
	// ivfvi => 5
	task.lengthOfLPS("vvvvivfvim");
	// civic => 5
	task.lengthOfLPS("xyzacivicxrs");
	// opo => 3
	task.lengthOfLPS("ttoopo");
}

Output

 Given string : nitiiiigi
 Resultant palindrome length : 4
 Given string : vvvvivfvim
 Resultant palindrome length : 5
 Given string : xyzacivicxrs
 Resultant palindrome length : 5
 Given string : ttoopo
 Resultant palindrome length : 3
Last updated on by Kalkicode




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