Longest common subsequence of three sequences

Here given code implementation process.

/*
    C program for
    Longest common subsequence of three sequences
*/
#include <stdio.h>
#include <string.h>

int maxValue(int a, int b)
{
	if (a > b)
	{
		return a;
	}
	return b;
}
void lengthOfLCP(char *text1, char *text2, char *text3)
{
	// Get the length of text1 and text2
	int m = strlen(text1);
	int n = strlen(text2);
	int o = strlen(text3);
	int dp[m][n][o];
	// Execute loop through by size of text1
	for (int i = 0; i < m; ++i)
	{
		// Execute loop through by size of text2
		for (int j = 0; j < n; ++j)
		{
			// Execute loop through by size of text3
			for (int k = 0; k < o; ++k)
			{
				if (i == 0 || j == 0 || k == 0)
				{
					dp[i][j][k] = 1;
				}
				else if (text1[i - 1] == text2[j - 1] && 
                         text2[j - 1] == text3[k - 1])
				{
					// When i-1 and j-1 and k-1 is character are same
					dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
				}
				else
				{
					// Get max value
					dp[i][j][k] = maxValue(
                      maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                      dp[i][j][k - 1]);
				}
			}
		}
	}
	// Given string
	printf("\n Given text1 : %s", text1);
	printf("\n Given text2 : %s", text2);
	printf("\n Given text3 : %s", text3);
	// Display length of LCS
	printf("\n Result : %d", dp[m - 1][n - 1][o - 1]);
}
int main(int argc, char
	const *argv[])
{
	char *text1 = "pdfpofindther";
	char *text2 = "pminoiwinters";
	char *text3 = "codepokingtenter";
	// "pointer" Common sequences
	// Result : 7
	lengthOfLCP(text1, text2, text3);
	return 0;
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
/*
    Java program for
    Longest common subsequence of three sequences
*/
public class LCP
{
    public int maxValue(int a, int b)
    {
        if (a > b)
        {
            return a;
        }
        return b;
    }
    public void lengthOfLCP(String text1, String text2, String text3)
    {
        // Get the length of text1 and text2
        int m = text1.length();
        int n = text2.length();
        int o = text3.length();
        int[][][] dp = new int[m][n][o];
        // Execute loop through by size of text1
        for (int i = 0; i < m; ++i)
        {
            // Execute loop through by size of text2
            for (int j = 0; j < n; ++j)
            {
                // Execute loop through by size of text3
                for (int k = 0; k < o; ++k)
                {
                    if (i == 0 || j == 0 || k == 0)
                    {
                        dp[i][j][k] = 1;
                    }
                    else if (text1.charAt(i - 1) == text2.charAt(j - 1) && 
                             text2.charAt(j - 1) == text3.charAt(k - 1))
                    {
                        // When i-1 and j-1 and k-1 is character are same
                        dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
                    }
                    else
                    {
                        // Get max value
                        dp[i][j][k] = maxValue(
                          maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1]);
                    }
                }
            }
        }
        // Given string
        System.out.print("\n Given text1 : " + text1);
        System.out.print("\n Given text2 : " + text2);
        System.out.print("\n Given text3 : " + text3);
        // Display length of LCS
        System.out.print("\n Result : " + dp[m - 1][n - 1][o - 1] );
    }
    public static void main(String[] args)
    {
        LCP task = new LCP();
        String text1 = "pdfpofindther";
        String text2 = "pminoiwinters";
        String text3 = "codepokingtenter";
        task.lengthOfLCP(text1, text2,text3);
    }
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ program for
    Longest common subsequence of three sequences
*/
class LCP
{
	public: int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	void lengthOfLCP(string text1, string text2, string text3)
	{
		// Get the length of text1 and text2
		int m = text1.length();
		int n = text2.length();
		int o = text3.length();
		int dp[m][n][o];
		// Execute loop through by size of text1
		for (int i = 0; i < m; ++i)
		{
			// Execute loop through by size of text2
			for (int j = 0; j < n; ++j)
			{
				// Execute loop through by size of text3
				for (int k = 0; k < o; ++k)
				{
					if (i == 0 || j == 0 || k == 0)
					{
						dp[i][j][k] = 1;
					}
					else if (text1[i - 1] == text2[j - 1] && 
                             text2[j - 1] == text3[k - 1])
					{
						// When i-1 and j-1 and k-1 is character are same
						dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
					}
					else
					{
						// Get max value
						dp[i][j][k] = this->maxValue(
                          this->maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1]);
					}
				}
			}
		}
		// Given string
		cout << "\n Given text1 : " << text1;
		cout << "\n Given text2 : " << text2;
		cout << "\n Given text3 : " << text3;
		// Display length of LCS
		cout << "\n Result : " << dp[m - 1][n - 1][o - 1];
	}
};
int main()
{
	LCP *task = new LCP();
	string text1 = "pdfpofindther";
	string text2 = "pminoiwinters";
	string text3 = "codepokingtenter";
	task->lengthOfLCP(text1, text2, text3);
	return 0;
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
// Include namespace system
using System;
/*
    Csharp program for
    Longest common subsequence of three sequences
*/
public class LCP
{
	public int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	public void lengthOfLCP(String text1, String text2, String text3)
	{
		// Get the length of text1 and text2
		int m = text1.Length;
		int n = text2.Length;
		int o = text3.Length;
		int[,,] dp = new int[m,n,o];
		// Execute loop through by size of text1
		for (int i = 0; i < m; ++i)
		{
			// Execute loop through by size of text2
			for (int j = 0; j < n; ++j)
			{
				// Execute loop through by size of text3
				for (int k = 0; k < o; ++k)
				{
					if (i == 0 || j == 0 || k == 0)
					{
						dp[i,j,k] = 1;
					}
					else if (text1[i - 1] == text2[j - 1] && 
                             text2[j - 1] == text3[k - 1])
					{
						// When i-1 and j-1 and k-1 is character are same
						dp[i,j,k] = dp[i - 1,j - 1,k - 1] + 1;
					}
					else
					{
						// Get max value
						dp[i,j,k] = this.maxValue(
                          this.maxValue(dp[i - 1,j,k], dp[i,j - 1,k]), 
                          dp[i,j,k - 1]);
					}
				}
			}
		}
		// Given string
		Console.Write("\n Given text1 : " + text1);
		Console.Write("\n Given text2 : " + text2);
		Console.Write("\n Given text3 : " + text3);
		// Display length of LCS
		Console.Write("\n Result : " + dp[m - 1,n - 1,o - 1]);
	}
	public static void Main(String[] args)
	{
		LCP task = new LCP();
		String text1 = "pdfpofindther";
		String text2 = "pminoiwinters";
		String text3 = "codepokingtenter";
		task.lengthOfLCP(text1, text2, text3);
	}
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
package main
import "fmt"
/*
    Go program for
    Longest common subsequence of three sequences
*/

func maxValue(a, b int) int {
	if a > b {
		return a
	}
	return b
}
func lengthOfLCP(text1, text2, text3 string) {
	// Get the length of text1 and text2
	var m int = len(text1)
	var n int = len(text2)
	var o int = len(text3)
	var dp = make([][][]int,m);
	for i := 0 ; i < m ; i++ {
		dp[i] = make([][]int,n)
		for j := 0 ; j < n ; j++ {
			dp[i][j] = make([]int,o)
		}
	}

	// Execute loop through by size of text1
	for i := 0 ; i < m ; i++ {
		// Execute loop through by size of text2
		for j := 0 ; j < n ; j++ {
			// Execute loop through by size of text3
			for k := 0 ; k < o ; k++ {
				if i == 0 || j == 0 || k == 0 {
					dp[i][j][k] = 1
				} else if text1[i - 1] == text2[j - 1] && text2[j - 1] == text3[k - 1] {
					// When i-1 and j-1 and k-1 is character are same
					dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1
				} else {
					// Get max value
					dp[i][j][k] = maxValue(maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), dp[i][j][k - 1])
				}
			}
		}
	}
	// Given string
	fmt.Print("\n Given text1 : ", text1)
	fmt.Print("\n Given text2 : ", text2)
	fmt.Print("\n Given text3 : ", text3)
	// Display length of LCS
	fmt.Print("\n Result : ", dp[m - 1][n - 1][o - 1])
}
func main() {

	var text1 string = "pdfpofindther"
	var text2 string = "pminoiwinters"
	var text3 string = "codepokingtenter"
	lengthOfLCP(text1, text2, text3)
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
<?php
/*
    Php program for
    Longest common subsequence of three sequences
*/
class LCP
{
	public	function maxValue($a, $b)
	{
		if ($a > $b)
		{
			return $a;
		}
		return $b;
	}
	public	function lengthOfLCP($text1, $text2, $text3)
	{
		// Get the length of text1 and text2
		$m = strlen($text1);
		$n = strlen($text2);
		$o = strlen($text3);
		$dp = array_fill(0, $m, array_fill(0, $n, array_fill(0, $o, 0)));
		// Execute loop through by size of text1
		for ($i = 0; $i < $m; ++$i)
		{
			// Execute loop through by size of text2
			for ($j = 0; $j < $n; ++$j)
			{
				// Execute loop through by size of text3
				for ($k = 0; $k < $o; ++$k)
				{
					if ($i == 0 || $j == 0 || $k == 0)
					{
						$dp[$i][$j][$k] = 1;
					}
					else if ($text1[$i - 1] == $text2[$j - 1] && 
                             $text2[$j - 1] == $text3[$k - 1])
					{
						// When i-1 and j-1 and k-1 is character are same
						$dp[$i][$j][$k] = $dp[$i - 1][$j - 1][$k - 1] + 1;
					}
					else
					{
						// Get max value
						$dp[$i][$j][$k] = $this->maxValue(
                       		$this->maxValue($dp[$i - 1][$j][$k], 
                                       $dp[$i][$j - 1][$k]), 
                          $dp[$i][$j][$k - 1]);
					}
				}
			}
		}
		// Given string
		echo("\n Given text1 : ".$text1);
		echo("\n Given text2 : ".$text2);
		echo("\n Given text3 : ".$text3);
		// Display length of LCS
		echo("\n Result : ".$dp[$m - 1][$n - 1][$o - 1]);
	}
}

function main()
{
	$task = new LCP();
	$text1 = "pdfpofindther";
	$text2 = "pminoiwinters";
	$text3 = "codepokingtenter";
	$task->lengthOfLCP($text1, $text2, $text3);
}
main();

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
/*
    Node JS program for
    Longest common subsequence of three sequences
*/
class LCP
{
	maxValue(a, b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	lengthOfLCP(text1, text2, text3)
	{
		// Get the length of text1 and text2
		var m = text1.length;
		var n = text2.length;
		var o = text3.length;
		var dp = 
            Array(m).fill(0).map(() => new Array(n).fill(0).map(
              () => new Array(o).fill(0)));
		// Execute loop through by size of text1
		for (var i = 0; i < m; ++i)
		{
			// Execute loop through by size of text2
			for (var j = 0; j < n; ++j)
			{
				// Execute loop through by size of text3
				for (var k = 0; k < o; ++k)
				{
					if (i == 0 || j == 0 || k == 0)
					{
						dp[i][j][k] = 1;
					}
					else if (text1.charAt(i - 1) == text2.charAt(j - 1) && 
                             text2.charAt(j - 1) == text3.charAt(k - 1))
					{
						// When i-1 and j-1 and k-1 is character are same
						dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
					}
					else
					{
						// Get max value
						dp[i][j][k] = this.maxValue(
                          this.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1]);
					}
				}
			}
		}
		// Given string
		process.stdout.write("\n Given text1 : " + text1);
		process.stdout.write("\n Given text2 : " + text2);
		process.stdout.write("\n Given text3 : " + text3);
		// Display length of LCS
		process.stdout.write("\n Result : " + dp[m - 1][n - 1][o - 1]);
	}
}

function main()
{
	var task = new LCP();
	var text1 = "pdfpofindther";
	var text2 = "pminoiwinters";
	var text3 = "codepokingtenter";
	task.lengthOfLCP(text1, text2, text3);
}
main();

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
#    Python 3 program for
#    Longest common subsequence of three sequences
class LCP :
	def maxValue(self, a, b) :
		if (a > b) :
			return a
		
		return b
	
	def lengthOfLCP(self, text1, text2, text3) :
		#  Get the length of text1 and text2
		m = len(text1)
		n = len(text2)
		o = len(text3)
		dp = [[[0] * (o) for _ in range(n) ]
		for _ in range(m) ]
		i = 0
		#  Execute loop through by size of text1
		while (i < m) :
			j = 0
			#  Execute loop through by size of text2
			while (j < n) :
				k = 0
				#  Execute loop through by size of text3
				while (k < o) :
					if (i == 0 or j == 0 or k == 0) :
						dp[i][j][k] = 1
					elif (text1[i - 1] == text2[j - 1] and 
                          text2[j - 1] == text3[k - 1]) :
						#  When i-1 and j-1 and k-1 is character are same
						dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1
					else :
						#  Get max value
						dp[i][j][k] = self.maxValue(
                          self.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1])
					
					k += 1
				
				j += 1
			
			i += 1
		
		#  Given string
		print("\n Given text1 : ", text1, end = "")
		print("\n Given text2 : ", text2, end = "")
		print("\n Given text3 : ", text3, end = "")
		#  Display length of LCS
		print("\n Result : ", dp[m - 1][n - 1][o - 1], end = "")
	

def main() :
	task = LCP()
	text1 = "pdfpofindther"
	text2 = "pminoiwinters"
	text3 = "codepokingtenter"
	task.lengthOfLCP(text1, text2, text3)

if __name__ == "__main__": main()

Output

 Given text1 :  pdfpofindther
 Given text2 :  pminoiwinters
 Given text3 :  codepokingtenter
 Result :  7
#    Ruby program for
#    Longest common subsequence of three sequences
class LCP 
	def maxValue(a, b) 
		if (a > b) 
			return a
		end

		return b
	end

	def lengthOfLCP(text1, text2, text3) 
		#  Get the length of text1 and text2
		m = text1.length
		n = text2.length
		o = text3.length
		dp = Array.new(m) {Array.new(n) {Array.new(o) {0}}}
		i = 0
		#  Execute loop through by size of text1
		while (i < m) 
			j = 0
			#  Execute loop through by size of text2
			while (j < n) 
				k = 0
				#  Execute loop through by size of text3
				while (k < o) 
					if (i == 0 || j == 0 || k == 0) 
						dp[i][j][k] = 1
					elsif (text1[i - 1] == text2[j - 1] && 
                           text2[j - 1] == text3[k - 1]) 
						#  When i-1 and j-1 and k-1 is character are same
						dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1
					else
 
						#  Get max value
						dp[i][j][k] = self.maxValue(
                          self.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1])
					end

					k += 1
				end

				j += 1
			end

			i += 1
		end

		#  Given string
		print("\n Given text1 : ", text1)
		print("\n Given text2 : ", text2)
		print("\n Given text3 : ", text3)
		#  Display length of LCS
		print("\n Result : ", dp[m - 1][n - 1][o - 1])
	end

end

def main() 
	task = LCP.new()
	text1 = "pdfpofindther"
	text2 = "pminoiwinters"
	text3 = "codepokingtenter"
	task.lengthOfLCP(text1, text2, text3)
end

main()

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
/*
    Scala program for
    Longest common subsequence of three sequences
*/
class LCP()
{
	def maxValue(a: Int, b: Int): Int = {
		if (a > b)
		{
			return a;
		}
		return b;
	}
	def lengthOfLCP(text1: String, text2: String, text3: String): Unit = {
		// Get the length of text1 and text2
		var m: Int = text1.length();
		var n: Int = text2.length();
		var o: Int = text3.length();
		var dp: Array[Array[Array[Int]]] = Array.fill[Int](m, n, o)(0);
		var i: Int = 0;
		// Execute loop through by size of text1
		while (i < m)
		{
			var j: Int = 0;
			// Execute loop through by size of text2
			while (j < n)
			{
				var k: Int = 0;
				// Execute loop through by size of text3
				while (k < o)
				{
					if (i == 0 || j == 0 || k == 0)
					{
						dp(i)(j)(k) = 1;
					}
					else if (text1.charAt(i - 1) == text2.charAt(j - 1) && 
                             text2.charAt(j - 1) == text3.charAt(k - 1))
					{
						// When i-1 and j-1 and k-1 is character are same
						dp(i)(j)(k) = dp(i - 1)(j - 1)(k - 1) + 1;
					}
					else
					{
						// Get max value
						dp(i)(j)(k) = maxValue(
                          maxValue(dp(i - 1)(j)(k), dp(i)(j - 1)(k)), 
                          dp(i)(j)(k - 1));
					}
					k += 1;
				}
				j += 1;
			}
			i += 1;
		}
		// Given string
		print("\n Given text1 : " + text1);
		print("\n Given text2 : " + text2);
		print("\n Given text3 : " + text3);
		// Display length of LCS
		print("\n Result : " + dp(m - 1)(n - 1)(o - 1));
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: LCP = new LCP();
		var text1: String = "pdfpofindther";
		var text2: String = "pminoiwinters";
		var text3: String = "codepokingtenter";
		task.lengthOfLCP(text1, text2, text3);
	}
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7
import Foundation;
/*
    Swift 4 program for
    Longest common subsequence of three sequences
*/
class LCP
{
	func maxValue(_ a: Int, _ b: Int) -> Int
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	func lengthOfLCP(_ a: String, _ b: String, _ c: String)
	{
		// Get the length of text1 and text2
		let m: Int = a.count;
		let n: Int = b.count;
		let o: Int = c.count;
      	let text1 = Array(a);
        let text2 = Array(b);
        let text3 = Array(c);
      
		var dp: [[[Int]]] = Array(
          repeating: Array(
          		repeating: Array(repeating: 0, count: o), count: n), count: m);
		var i: Int = 0;
		// Execute loop through by size of text1
		while (i < m)
		{
			var j: Int = 0;
			// Execute loop through by size of text2
			while (j < n)
			{
				var k: Int = 0;
				// Execute loop through by size of text3
				while (k < o)
				{
					if (i == 0 || j == 0 || k == 0)
					{
						dp[i][j][k] = 1;
					}
					else if (text1[i - 1] == text2[j - 1] && 
                             text2[j - 1] == text3[k - 1])
					{
						// When i-1 and j-1 and k-1 is character are same
						dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
					}
					else
					{
						// Get max value
						dp[i][j][k] = self.maxValue(
                          self.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1]);
					}
					k += 1;
				}
				j += 1;
			}
			i += 1;
		}
		// Given string
		print("\n Given text1 : ", a, terminator: "");
		print("\n Given text2 : ", b, terminator: "");
		print("\n Given text3 : ", c, terminator: "");
		// Display length of LCS
		print("\n Result : ", dp[m - 1][n - 1][o - 1], terminator: "");
	}
}
func main()
{
	let task: LCP = LCP();
	let text1: String = "pdfpofindther";
	let text2: String = "pminoiwinters";
	let text3: String = "codepokingtenter";
	task.lengthOfLCP(text1, text2, text3);
}
main();

Output

 Given text1 :  pdfpofindther
 Given text2 :  pminoiwinters
 Given text3 :  codepokingtenter
 Result :  7
/*
    Kotlin program for
    Longest common subsequence of three sequences
*/
class LCP
{
	fun maxValue(a: Int, b: Int): Int
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	fun lengthOfLCP(text1: String, text2: String, text3: String): Unit
	{
		// Get the length of text1 and text2
		val m: Int = text1.length;
		val n: Int = text2.length;
		val o: Int = text3.length;
		var dp: Array < Array < Array < Int >>> = Array(m)
		{
			Array(n)
			{
				Array(o)
				{
					0
				}
			}
		};
		var i: Int = 0;
		// Execute loop through by size of text1
		while (i < m)
		{
			var j: Int = 0;
			// Execute loop through by size of text2
			while (j < n)
			{
				var k: Int = 0;
				// Execute loop through by size of text3
				while (k < o)
				{
					if (i == 0 || j == 0 || k == 0)
					{
						dp[i][j][k] = 1;
					}
					else if (text1.get(i - 1) == text2.get(j - 1) && 
                             text2.get(j - 1) == text3.get(k - 1))
					{
						// When i-1 and j-1 and k-1 is character are same
						dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
					}
					else
					{
						// Get max value
						dp[i][j][k] = this.maxValue(
                          this.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), 
                          dp[i][j][k - 1]);
					}
					k += 1;
				}
				j += 1;
			}
			i += 1;
		}
		// Given string
		print("\n Given text1 : " + text1);
		print("\n Given text2 : " + text2);
		print("\n Given text3 : " + text3);
		// Display length of LCS
		print("\n Result : " + dp[m - 1][n - 1][o - 1]);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: LCP = LCP();
	val text1: String = "pdfpofindther";
	val text2: String = "pminoiwinters";
	val text3: String = "codepokingtenter";
	task.lengthOfLCP(text1, text2, text3);
}

Output

 Given text1 : pdfpofindther
 Given text2 : pminoiwinters
 Given text3 : codepokingtenter
 Result : 7


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