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Printing Shortest Common Supersequence

The problem is a classic algorithmic challenge that involves finding the shortest string that contains two given strings as subsequences. This problem is relevant in various areas such as DNA sequence alignment, text processing, and computational biology. In this article, we'll delve into the approach to solving the Shortest Common Supersequence problem using dynamic programming and provide a detailed explanation along with code examples.

Problem Statement and Description

Given two strings A and B, the goal of the Shortest Common Supersequence problem is to find the shortest string C that contains both A and B as subsequences. In other words, C should have A and B as subsequences while minimizing its length. The problem finds applications in tasks such as DNA sequence alignment, where researchers want to find the shortest sequence that includes both genetic sequences.

Example

Consider the following examples to grasp the concept:

  • For strings "abc" and "fab": The shortest common supersequence is "fabc", with a length of 4 characters.

  • For strings "project" and "objects": The shortest common supersequence is "probjects", with a length of 9 characters.

  • For strings "match" and "attack": One possible shortest common supersequence is "mattackh", with a length of 8 characters.

  • For strings "abc" and "abc": The shortest common supersequence is simply "abc", as it includes both strings and has a length of 3 characters.

Idea to Solve the Problem

To solve the Shortest Common Supersequence problem, we can utilize dynamic programming. The idea is to build a 2D table where each entry dp[i][j] represents the length of the shortest common supersequence of substrings A[0...i-1] and B[0...j-1].

Standard Pseudocode

function findSCS(a, b):
    n = length of string a
    m = length of string b
    dp = 2D array of size (n+1) x (m+1)
    result = empty string
    
    for i from 0 to n:
        for j from 0 to m:
            if i is 0:
                dp[i][j] = j
            else if j is 0:
                dp[i][j] = i
            else if a[i-1] is equal to b[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = minimum of (dp[i][j-1], dp[i-1][j]) + 1
    
    s = n
    e = m
    
    while s > 0 and e > 0:
        if a[s-1] is equal to b[e-1]:
            add a[s-1] to the beginning of result
            decrement s and e
        else if dp[s-1][e] > dp[s][e-1]:
            add b[e-1] to the beginning of result
            decrement e
        else:
            add a[s-1] to the beginning of result
            decrement s
    
    while s > 0:
        add a[s-1] to the beginning of result
        decrement s
    
    while e > 0:
        add b[e-1] to the beginning of result
        decrement e
    
    print "Given string a :", a
    print "Given string b :", b
    print "Length :", dp[n][m]
    print "Result :", result

Algorithm Explanation

  1. Initialize a 2D array dp of size (n+1) x (m+1) to store the lengths of the shortest common supersequences.
  2. Iterate through the arrays a and b using two nested loops, filling the dp array based on certain conditions.
  3. Traverse the dp array to reconstruct the shortest common supersequence and store it in the result variable.
  4. Print the original strings, the length of the shortest common supersequence, and the reconstructed supersequence.

Code Solution

/*
    Java program for
    Printing Shortest Common Supersequence
*/
public class Supersequence
{
	// Returns minimum of given values
	public int minValue(int x, int y)
	{
		if (x < y)
		{
			return x;
		}
		return y;
	}
	// Find length of shortest common Supersequence 
	public void findSCS(String a, String b)
	{
		int n = a.length();
		int m = b.length();
		String result = "";
		// Auxiliary space
		int[][] dp = new int[n + 1][m + 1];
		// Outer loop, executing this from 0 to n (length of a)
		for (int i = 0; i <= n; ++i)
		{
			// Inner loop, executing this from 0 to m (length of b)
			for (int j = 0; j <= m; ++j)
			{
				if (i == 0)
				{
					// When i is zero
					dp[i][j] = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp[i][j] = i;
				}
				else if (a.charAt(i - 1) == b.charAt(j - 1))
				{
					//  When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp[i][j] = minValue(dp[i][j - 1], dp[i - 1][j]) + 1;
				}
			}
		}
		// Display given strings
		System.out.println(" Given string a : " + a);
		System.out.println(" Given string b : " + b);
		int min = dp[n][m];
		int s = n;
		int e = m;
		while (s > 0 && e > 0)
		{
			if (a.charAt(s - 1) == b.charAt(e - 1))
			{
				// When both string character at position are same from end
				result = a.charAt(s - 1) + result;
				s--;
				e--;
			}
			else if (dp[s - 1][e] > dp[s][e - 1])
			{
				result = b.charAt(e - 1) + result;
				e--;
			}
			else
			{
				result = a.charAt(s - 1) + result;
				s--;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = a.charAt(s - 1) + result;
			s--;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = b.charAt(e - 1) + result;
			e--;
		}
		// Display calculated result
		System.out.println(" Length : " + dp[n][m]);
		System.out.println(" Result : " + result);
	}
	public static void main(String[] args)
	{
		Supersequence task = new Supersequence();
		// Test A
		// String a : abc
		// String b : fab
		// [fabc] Supersequence 
		// Result = 4
		task.findSCS("abc", "fab");
		// Test B
		// String a : project
		// String b : objects
		// [probjects]  
		// Result : 9
		task.findSCS("project", "objects");
		// Test C
		// String a : match
		// String b : attack
		// [mattachk,matatchk,mattackh] etc
		// Result : 8 (length of Supersequence)
		task.findSCS("match", "attack");
		// Test D
		// String a : abc
		// String b : abc
		// [abc] etc
		// Result : 3 
		task.findSCS("abc", "abc");
	}
}

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ program for
    Printing Shortest Common Supersequence
*/
class Supersequence
{
	public:
		// Returns minimum of given values
		int minValue(int x, int y)
		{
			if (x < y)
			{
				return x;
			}
			return y;
		}
	// Find length of shortest common Supersequence 
	void findSCS(string a, string b)
	{
		int n = a.length();
		int m = b.length();
		string result = "";
		// Auxiliary space
		int dp[n + 1][m + 1];
		// Outer loop, executing this from 0 to n (length of a)
		for (int i = 0; i <= n; ++i)
		{
			// Inner loop, executing this from 0 to m (length of b)
			for (int j = 0; j <= m; ++j)
			{
				if (i == 0)
				{
					// When i is zero
					dp[i][j] = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp[i][j] = i;
				}
				else if (a[i - 1] == b[j - 1])
				{
					//  When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp[i][j] = this->minValue(dp[i][j - 1], 
                                              dp[i - 1][j]) + 1;
				}
			}
		}
		// Display given strings
		cout << " Given string a : " << a << endl;
		cout << " Given string b : " << b << endl;
		int min = dp[n][m];
		int s = n;
		int e = m;
		while (s > 0 && e > 0)
		{
			if (a[s - 1] == b[e - 1])
			{
				// When both string character at position are same from end
				result = (a[s - 1])  +  result;
				s--;
				e--;
			}
			else if (dp[s - 1][e] > dp[s][e - 1])
			{
				result = (b[e - 1])  +  result;
				e--;
			}
			else
			{
				result = (a[s - 1])  +  result;
				s--;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = (a[s - 1])  +  result;
			s--;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = (b[e - 1])  +  result;
			e--;
		}
		// Display calculated result
		cout << " Length : " << dp[n][m] << endl;
		cout << " Result : " << result << endl;
	}
};
int main()
{
	Supersequence *task = new Supersequence();
	// Test A
	// String a : abc
	// String b : fab
	// [fabc] Supersequence 
	// Result = 4
	task->findSCS("abc", "fab");
	// Test B
	// String a : project
	// String b : objects
	// [probjects]  
	// Result : 9
	task->findSCS("project", "objects");
	// Test C
	// String a : match
	// String b : attack
	// [mattachk,matatchk,mattackh] etc
	// Result : 8 (length of Supersequence)
	task->findSCS("match", "attack");
	// Test D
	// String a : abc
	// String b : abc
	// [abc] etc
	// Result : 3 
	task->findSCS("abc", "abc");
	return 0;
}

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
// Include namespace system
using System;
/*
    Csharp program for
    Printing Shortest Common Supersequence
*/
public class Supersequence
{
	// Returns minimum of given values
	public int minValue(int x, int y)
	{
		if (x < y)
		{
			return x;
		}
		return y;
	}
	// Find length of shortest common Supersequence 
	public void findSCS(String a, String b)
	{
		int n = a.Length;
		int m = b.Length;
		String result = "";
		// Auxiliary space
		int[,] dp = new int[n + 1,m + 1];
		// Outer loop, executing this from 0 to n (length of a)
		for (int i = 0; i <= n; ++i)
		{
			// Inner loop, executing this from 0 to m (length of b)
			for (int j = 0; j <= m; ++j)
			{
				if (i == 0)
				{
					// When i is zero
					dp[i,j] = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp[i,j] = i;
				}
				else if (a[i - 1] == b[j - 1])
				{
					//  When character of i-1 and j-1 position are same
					dp[i,j] = dp[i - 1,j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp[i,j] = this.minValue(dp[i,j - 1], dp[i - 1,j]) + 1;
				}
			}
		}
		// Display given strings
		Console.WriteLine(" Given string a : " + a);
		Console.WriteLine(" Given string b : " + b);
		int min = dp[n,m];
		int s = n;
		int e = m;
		while (s > 0 && e > 0)
		{
			if (a[s - 1] == b[e - 1])
			{
				// When both string character at position are same from end
				result = a[s - 1] + result;
				s--;
				e--;
			}
			else if (dp[s - 1,e] > dp[s,e - 1])
			{
				result = b[e - 1] + result;
				e--;
			}
			else
			{
				result = a[s - 1] + result;
				s--;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = a[s - 1] + result;
			s--;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = b[e - 1] + result;
			e--;
		}
		// Display calculated result
		Console.WriteLine(" Length : " + min);
		Console.WriteLine(" Result : " + result);
	}
	public static void Main(String[] args)
	{
		Supersequence task = new Supersequence();
		// Test A
		// String a : abc
		// String b : fab
		// [fabc] Supersequence 
		// Result = 4
		task.findSCS("abc", "fab");
		// Test B
		// String a : project
		// String b : objects
		// [probjects]  
		// Result : 9
		task.findSCS("project", "objects");
		// Test C
		// String a : match
		// String b : attack
		// [mattachk,matatchk,mattackh] etc
		// Result : 8 (length of Supersequence)
		task.findSCS("match", "attack");
		// Test D
		// String a : abc
		// String b : abc
		// [abc] etc
		// Result : 3 
		task.findSCS("abc", "abc");
	}
}

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
package main
import "fmt"
/*
    Go program for
    Printing Shortest Common Supersequence
*/

// Returns minimum of given values
func minValue(x, y int) int {
	if x < y {
		return x
	}
	return y
}
// Find length of shortest common Supersequence 
func findSCS(a, b string) {
	var n int = len(a)
	var m int = len(b)
	var result string = ""
	// Auxiliary space
	var dp = make([][] int, n + 1)
	for i := 0; i < n + 1; i++ {
		dp[i] = make([] int, m + 1)
	}
	// Outer loop, executing this from 0 to n (length of a)
	for i := 0 ; i <= n ; i++ {
		// Inner loop, executing this from 0 to m (length of b)
		for j := 0 ; j <= m ; j++ {
			if i == 0 {
				// When i is zero
				dp[i][j] = j
			} else if j == 0 {
				// When j is zero
				dp[i][j] = i
			} else if a[i - 1] == b[j - 1] {
				//  When character of i-1 and j-1 position are same
				dp[i][j] = dp[i - 1][j - 1] + 1
			} else {
				//  When character of i-1 and j-1 position are not same
				dp[i][j] = minValue(dp[i][j - 1], dp[i - 1][j]) + 1
			}
		}
	}
	// Display given strings
	fmt.Println(" Given string a : ", a)
	fmt.Println(" Given string b : ", b)
	var min int = dp[n][m]
	var s int = n
	var e int = m
	for (s > 0 && e > 0) {
		if a[s - 1] == b[e - 1] {
			// When both string character at position are same from end
			result = string(a[s - 1]) + result
			s--
			e--
		} else if dp[s - 1][e] > dp[s][e - 1] {
			result = string(b[e - 1]) + result
			e--
		} else {
			result = string(a[s - 1]) + result
			s--
		}
	}
	// Collect remaining character of first string
	for (s > 0) {
		result = string(a[s - 1]) + result
		s--
	}
	// Collect remaining character of second string 
	for (e > 0) {
		result = string(b[e - 1]) + result
		e--
	}
	// Display calculated result
	fmt.Println(" Length : ", min)
	fmt.Println(" Result : ", result)
}
func main() {

	// Test A
	// String a : abc
	// String b : fab
	// [fabc] Supersequence 
	// Result = 4
	findSCS("abc", "fab")
	// Test B
	// String a : project
	// String b : objects
	// [probjects]  
	// Result : 9
	findSCS("project", "objects")
	// Test C
	// String a : match
	// String b : attack
	// [mattachk,matatchk,mattackh] etc
	// Result : 8 (length of Supersequence)
	findSCS("match", "attack")
	// Test D
	// String a : abc
	// String b : abc
	// [abc] etc
	// Result : 3 
	findSCS("abc", "abc")
}

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
<?php
/*
    Php program for
    Printing Shortest Common Supersequence
*/
class Supersequence
{
	// Returns minimum of given values
	public	function minValue($x, $y)
	{
		if ($x < $y)
		{
			return $x;
		}
		return $y;
	}
	// Find length of shortest common Supersequence 
	public	function findSCS($a, $b)
	{
		$n = strlen($a);
		$m = strlen($b);
		$result = "";
		// Auxiliary space
		$dp = array_fill(0, $n + 1, array_fill(0, $m + 1, 0));
		// Outer loop, executing this from 0 to n (length of a)
		for ($i = 0; $i <= $n; ++$i)
		{
			// Inner loop, executing this from 0 to m (length of b)
			for ($j = 0; $j <= $m; ++$j)
			{
				if ($i == 0)
				{
					// When i is zero
					$dp[$i][$j] = $j;
				}
				else if ($j == 0)
				{
					// When j is zero
					$dp[$i][$j] = $i;
				}
				else if ($a[$i - 1] == $b[$j - 1])
				{
					//  When character of i-1 and j-1 position are same
					$dp[$i][$j] = $dp[$i - 1][$j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					$dp[$i][$j] = $this->minValue(
                      $dp[$i][$j - 1], $dp[$i - 1][$j]) + 1;
				}
			}
		}
		// Display given strings
		echo(" Given string a : ".$a.
			"\n");
		echo(" Given string b : ".$b.
			"\n");
		$min = $dp[$n][$m];
		$s = $n;
		$e = $m;
		while ($s > 0 && $e > 0)
		{
			if ($a[$s - 1] == $b[$e - 1])
			{
				// When both string character at position are same from end
				$result = strval($a[$s - 1]).$result;
				$s--;
				$e--;
			}
			else if ($dp[$s - 1][$e] > $dp[$s][$e - 1])
			{
				$result = strval($b[$e - 1]).$result;
				$e--;
			}
			else
			{
				$result = strval($a[$s - 1]).$result;
				$s--;
			}
		}
		// Collect remaining character of first string
		while ($s > 0)
		{
			$result = strval($a[$s - 1]).$result;
			$s--;
		}
		// Collect remaining character of second string 
		while ($e > 0)
		{
			$result = strval($b[$e - 1]).$result;
			$e--;
		}
		// Display calculated result
		echo(" Length : ".$min."\n");
		echo(" Result : ".$result."\n");
	}
}

function main()
{
	$task = new Supersequence();
	// Test A
	// String a : abc
	// String b : fab
	// [fabc] Supersequence 
	// Result = 4
	$task->findSCS("abc", "fab");
	// Test B
	// String a : project
	// String b : objects
	// [probjects]  
	// Result : 9
	$task->findSCS("project", "objects");
	// Test C
	// String a : match
	// String b : attack
	// [mattachk,matatchk,mattackh] etc
	// Result : 8 (length of Supersequence)
	$task->findSCS("match", "attack");
	// Test D
	// String a : abc
	// String b : abc
	// [abc] etc
	// Result : 3 
	$task->findSCS("abc", "abc");
}
main();

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
/*
    Node JS program for
    Printing Shortest Common Supersequence
*/
class Supersequence
{
	// Returns minimum of given values
	minValue(x, y)
	{
		if (x < y)
		{
			return x;
		}
		return y;
	}
	// Find length of shortest common Supersequence 
	findSCS(a, b)
	{
		var n = a.length;
		var m = b.length;
		var result = "";
		// Auxiliary space
		var dp = Array(n + 1).fill(0).map(() => new Array(m + 1).fill(0));
		// Outer loop, executing this from 0 to n (length of a)
		for (var i = 0; i <= n; ++i)
		{
			// Inner loop, executing this from 0 to m (length of b)
			for (var j = 0; j <= m; ++j)
			{
				if (i == 0)
				{
					// When i is zero
					dp[i][j] = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp[i][j] = i;
				}
				else if (a.charAt(i - 1) == b.charAt(j - 1))
				{
					//  When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp[i][j] = this.minValue(dp[i][j - 1], dp[i - 1][j]) + 1;
				}
			}
		}
		// Display given strings
		console.log(" Given string a : " + a);
		console.log(" Given string b : " + b);
		var min = dp[n][m];
		var s = n;
		var e = m;
		while (s > 0 && e > 0)
		{
			if (a.charAt(s - 1) == b.charAt(e - 1))
			{
				// When both string character at position are same from end
				result = a.charAt(s - 1) + result;
				s--;
				e--;
			}
			else if (dp[s - 1][e] > dp[s][e - 1])
			{
				result = b.charAt(e - 1) + result;
				e--;
			}
			else
			{
				result = a.charAt(s - 1) + result;
				s--;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = a.charAt(s - 1) + result;
			s--;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = b.charAt(e - 1) + result;
			e--;
		}
		// Display calculated result
		console.log(" Length : " + min);
		console.log(" Result : " + result);
	}
}

function main()
{
	var task = new Supersequence();
	// Test A
	// String a : abc
	// String b : fab
	// [fabc] Supersequence 
	// Result = 4
	task.findSCS("abc", "fab");
	// Test B
	// String a : project
	// String b : objects
	// [probjects]  
	// Result : 9
	task.findSCS("project", "objects");
	// Test C
	// String a : match
	// String b : attack
	// [mattachk,matatchk,mattackh] etc
	// Result : 8 (length of Supersequence)
	task.findSCS("match", "attack");
	// Test D
	// String a : abc
	// String b : abc
	// [abc] etc
	// Result : 3 
	task.findSCS("abc", "abc");
}
main();

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
#    Python 3 program for
#    Printing Shortest Common Supersequence
class Supersequence :
	#  Returns minimum of given values
	def minValue(self, x, y) :
		if (x < y) :
			return x
		
		return y
	
	#  Find length of shortest common Supersequence 
	def findSCS(self, a, b) :
		n = len(a)
		m = len(b)
		result = ""
		#  Auxiliary space
		dp = [[0] * (m + 1) for _ in range(n + 1) ]
		i = 0
		#  Outer loop, executing this from 0 to n (length of a)
		while (i <= n) :
			j = 0
			#  Inner loop, executing this from 0 to m (length of b)
			while (j <= m) :
				if (i == 0) :
					#  When i is zero
					dp[i][j] = j
				elif (j == 0) :
					#  When j is zero
					dp[i][j] = i
				elif (a[i - 1] == b[j - 1]) :
					#   When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1
				else :
					#   When character of i-1 and j-1 position are not same
					dp[i][j] = self.minValue(dp[i][j - 1], dp[i - 1][j]) + 1
				
				j += 1
			
			i += 1
		
		#  Display given strings
		print(" Given string a : ", a)
		print(" Given string b : ", b)
		min = dp[n][m]
		s = n
		e = m
		while (s > 0 and e > 0) :
			if (a[s - 1] == b[e - 1]) :
				#  When both string character at position are same from end
				result = str(a[s - 1]) + result
				s -= 1
				e -= 1
			elif (dp[s - 1][e] > dp[s][e - 1]) :
				result = str(b[e - 1]) + result
				e -= 1
			else :
				result = str(a[s - 1]) + result
				s -= 1
			
		
		#  Collect remaining character of first string
		while (s > 0) :
			result = str(a[s - 1]) + result
			s -= 1
		
		#  Collect remaining character of second string 
		while (e > 0) :
			result = str(b[e - 1]) + result
			e -= 1
		
		#  Display calculated result
		print(" Length : ", min)
		print(" Result : ", result)
	

def main() :
	task = Supersequence()
	#  Test A
	#  String a : abc
	#  String b : fab
	#  [fabc] Supersequence 
	#  Result = 4
	task.findSCS("abc", "fab")
	#  Test B
	#  String a : project
	#  String b : objects
	#  [probjects]  
	#  Result : 9
	task.findSCS("project", "objects")
	#  Test C
	#  String a : match
	#  String b : attack
	#  [mattachk,matatchk,mattackh] etc
	#  Result : 8 (length of Supersequence)
	task.findSCS("match", "attack")
	#  Test D
	#  String a : abc
	#  String b : abc
	#  [abc] etc
	#  Result : 3 
	task.findSCS("abc", "abc")

if __name__ == "__main__": main()

Output

 Given string a :  abc
 Given string b :  fab
 Length :  4
 Result :  fabc
 Given string a :  project
 Given string b :  objects
 Length :  9
 Result :  probjects
 Given string a :  match
 Given string b :  attack
 Length :  8
 Result :  mattackh
 Given string a :  abc
 Given string b :  abc
 Length :  3
 Result :  abc
#    Ruby program for
#    Printing Shortest Common Supersequence
class Supersequence 
	#  Returns minimum of given values
	def minValue(x, y) 
		if (x < y) 
			return x
		end

		return y
	end

	#  Find length of shortest common Supersequence 
	def findSCS(a, b) 
		n = a.length
		m = b.length
		result = ""
		#  Auxiliary space
		dp = Array.new(n + 1) {Array.new(m + 1) {0}}
		i = 0
		#  Outer loop, executing this from 0 to n (length of a)
		while (i <= n) 
			j = 0
			#  Inner loop, executing this from 0 to m (length of b)
			while (j <= m) 
				if (i == 0) 
					#  When i is zero
					dp[i][j] = j
				elsif (j == 0) 
					#  When j is zero
					dp[i][j] = i
				elsif (a[i - 1] == b[j - 1]) 
					#   When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1
				else
 
					#   When character of i-1 and j-1 position are not same
					dp[i][j] = self.minValue(dp[i][j - 1], dp[i - 1][j]) + 1
				end

				j += 1
			end

			i += 1
		end

		#  Display given strings
		print(" Given string a : ", a, "\n")
		print(" Given string b : ", b, "\n")
		min = dp[n][m]
		s = n
		e = m
		while (s > 0 && e > 0) 
			if (a[s - 1] == b[e - 1]) 
				#  When both string character at position are same from end
				result = a[s - 1].to_s + result
				s -= 1
				e -= 1
			elsif (dp[s - 1][e] > dp[s][e - 1]) 
				result = b[e - 1].to_s + result
				e -= 1
			else
 
				result = a[s - 1].to_s + result
				s -= 1
			end

		end

		#  Collect remaining character of first string
		while (s > 0) 
			result = a[s - 1].to_s + result
			s -= 1
		end

		#  Collect remaining character of second string 
		while (e > 0) 
			result = b[e - 1].to_s + result
			e -= 1
		end

		#  Display calculated result
		print(" Length : ", min, "\n")
		print(" Result : ", result, "\n")
	end

end

def main() 
	task = Supersequence.new()
	#  Test A
	#  String a : abc
	#  String b : fab
	#  [fabc] Supersequence 
	#  Result = 4
	task.findSCS("abc", "fab")
	#  Test B
	#  String a : project
	#  String b : objects
	#  [probjects]  
	#  Result : 9
	task.findSCS("project", "objects")
	#  Test C
	#  String a : match
	#  String b : attack
	#  [mattachk,matatchk,mattackh] etc
	#  Result : 8 (length of Supersequence)
	task.findSCS("match", "attack")
	#  Test D
	#  String a : abc
	#  String b : abc
	#  [abc] etc
	#  Result : 3 
	task.findSCS("abc", "abc")
end

main()

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc

import scala.collection.mutable._;
/*
    Scala program for
    Printing Shortest Common Supersequence
*/
class Supersequence()
{
	// Returns minimum of given values
	def minValue(x: Int, y: Int): Int = {
		if (x < y)
		{
			return x;
		}
		return y;
	}
	// Find length of shortest common Supersequence 
	def findSCS(a: String, b: String): Unit = {
		var n: Int = a.length();
		var m: Int = b.length();
		var result: String = "";
		// Auxiliary space
		var dp: Array[Array[Int]] = Array.fill[Int](n + 1, m + 1)(0);
		var i: Int = 0;
		// Outer loop, executing this from 0 to n (length of a)
		while (i <= n)
		{
			var j: Int = 0;
			// Inner loop, executing this from 0 to m (length of b)
			while (j <= m)
			{
				if (i == 0)
				{
					// When i is zero
					dp(i)(j) = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp(i)(j) = i;
				}
				else if (a.charAt(i - 1) == b.charAt(j - 1))
				{
					//  When character of i-1 and j-1 position are same
					dp(i)(j) = dp(i - 1)(j - 1) + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp(i)(j) = minValue(dp(i)(j - 1), dp(i - 1)(j)) + 1;
				}
				j += 1;
			}
			i += 1;
		}
		// Display given strings
		println(" Given string a : " + a);
		println(" Given string b : " + b);
		var min: Int = dp(n)(m);
		var s: Int = n;
		var e: Int = m;
		while (s > 0 && e > 0)
		{
			if (a.charAt(s - 1) == b.charAt(e - 1))
			{
				// When both string character at position are same from end
				result = a.charAt(s - 1).toString() + result;
				s -= 1;
				e -= 1;
			}
			else if (dp(s - 1)(e) > dp(s)(e - 1))
			{
				result = b.charAt(e - 1).toString() + result;
				e -= 1;
			}
			else
			{
				result = a.charAt(s - 1).toString() + result;
				s -= 1;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = a.charAt(s - 1).toString() + result;
			s -= 1;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = b.charAt(e - 1).toString() + result;
			e -= 1;
		}
		// Display calculated result
		println(" Length : " + min);
		println(" Result : " + result);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Supersequence = new Supersequence();
		// Test A
		// String a : abc
		// String b : fab
		// [fabc] Supersequence 
		// Result = 4
		task.findSCS("abc", "fab");
		// Test B
		// String a : project
		// String b : objects
		// [probjects]  
		// Result : 9
		task.findSCS("project", "objects");
		// Test C
		// String a : match
		// String b : attack
		// [mattachk,matatchk,mattackh] etc
		// Result : 8 (length of Supersequence)
		task.findSCS("match", "attack");
		// Test D
		// String a : abc
		// String b : abc
		// [abc] etc
		// Result : 3 
		task.findSCS("abc", "abc");
	}
}

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc
import Foundation;
/*
    Swift 4 program for
    Printing Shortest Common Supersequence
*/
class Supersequence
{
	// Returns minimum of given values
	func minValue(_ x: Int, _ y: Int) -> Int
	{
		if (x < y)
		{
			return x;
		}
		return y;
	}
	// Find length of shortest common Supersequence 
	func findSCS(_ a1: String, _ b1: String)
	{
      	let a = Array(a1);
      	let b = Array(b1);
		let n: Int = a.count;
		let m: Int = b.count;
		var result: String = "";
		// Auxiliary space
		var dp: [
			[Int]
		] = Array(
          repeating: Array(repeating: 0, count: m + 1), 
          count: n + 1);
		var i: Int = 0;
		// Outer loop, executing this from 0 to n (length of a)
		while (i <= n)
		{
			var j: Int = 0;
			// Inner loop, executing this from 0 to m (length of b)
			while (j <= m)
			{
				if (i == 0)
				{
					// When i is zero
					dp[i][j] = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp[i][j] = i;
				}
				else if (a[i - 1] == b[j - 1])
				{
					//  When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp[i][j] = self.minValue(dp[i][j - 1], dp[i - 1][j]) + 1;
				}
				j += 1;
			}
			i += 1;
		}
		// Display given strings
		print(" Given string a : ", a1);
		print(" Given string b : ", b1);
		let min: Int = dp[n][m];
		var s: Int = n;
		var e: Int = m;
		while (s > 0 && e > 0)
		{
			if (a[s - 1] == b[e - 1])
			{
				// When both string character at position are same from end
				result = String(a[s - 1]) + result;
				s -= 1;
				e -= 1;
			}
			else if (dp[s - 1][e] > dp[s][e - 1])
			{
				result = String(b[e - 1]) + result;
				e -= 1;
			}
			else
			{
				result = String(a[s - 1]) + result;
				s -= 1;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = String(a[s - 1]) + result;
			s -= 1;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = String(b[e - 1]) + result;
			e -= 1;
		}
		// Display calculated result
		print(" Length : ", min);
		print(" Result : ", result);
	}
}
func main()
{
	let task: Supersequence = Supersequence();
	// Test A
	// String a : abc
	// String b : fab
	// [fabc] Supersequence 
	// Result = 4
	task.findSCS("abc", "fab");
	// Test B
	// String a : project
	// String b : objects
	// [probjects]  
	// Result : 9
	task.findSCS("project", "objects");
	// Test C
	// String a : match
	// String b : attack
	// [mattachk,matatchk,mattackh] etc
	// Result : 8 (length of Supersequence)
	task.findSCS("match", "attack");
	// Test D
	// String a : abc
	// String b : abc
	// [abc] etc
	// Result : 3 
	task.findSCS("abc", "abc");
}
main();

Output

 Given string a :  abc
 Given string b :  fab
 Length :  4
 Result :  fabc
 Given string a :  project
 Given string b :  objects
 Length :  9
 Result :  probjects
 Given string a :  match
 Given string b :  attack
 Length :  8
 Result :  mattackh
 Given string a :  abc
 Given string b :  abc
 Length :  3
 Result :  abc
/*
    Kotlin program for
    Printing Shortest Common Supersequence
*/
class Supersequence
{
	// Returns minimum of given values
	fun minValue(x: Int, y: Int): Int
	{
		if (x < y)
		{
			return x;
		}
		return y;
	}
	// Find length of shortest common Supersequence 
	fun findSCS(a: String, b: String): Unit
	{
		val n: Int = a.length;
		val m: Int = b.length;
		var result: String = "";
		// Auxiliary space
		var dp: Array < Array < Int >> = Array(n + 1)
		{
			Array(m + 1)
			{
				0
			}
		};
		var i: Int = 0;
		// Outer loop, executing this from 0 to n (length of a)
		while (i <= n)
		{
			var j: Int = 0;
			// Inner loop, executing this from 0 to m (length of b)
			while (j <= m)
			{
				if (i == 0)
				{
					// When i is zero
					dp[i][j] = j;
				}
				else if (j == 0)
				{
					// When j is zero
					dp[i][j] = i;
				}
				else if (a.get(i - 1) == b.get(j - 1))
				{
					//  When character of i-1 and j-1 position are same
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else
				{
					//  When character of i-1 and j-1 position are not same
					dp[i][j] = this.minValue(dp[i][j - 1], dp[i - 1][j]) + 1;
				}
				j += 1;
			}
			i += 1;
		}
		// Display given strings
		println(" Given string a : " + a);
		println(" Given string b : " + b);
		val min: Int = dp[n][m];
		var s: Int = n;
		var e: Int = m;
		while (s > 0 && e > 0)
		{
			if (a.get(s - 1) == b.get(e - 1))
			{
				// When both string character at position are same from end
				result = a.get(s - 1).toString() + result;
				s -= 1;
				e -= 1;
			}
			else if (dp[s - 1][e] > dp[s][e - 1])
			{
				result = b.get(e - 1).toString() + result;
				e -= 1;
			}
			else
			{
				result = a.get(s - 1).toString() + result;
				s -= 1;
			}
		}
		// Collect remaining character of first string
		while (s > 0)
		{
			result = a.get(s - 1).toString() + result;
			s -= 1;
		}
		// Collect remaining character of second string 
		while (e > 0)
		{
			result = b.get(e - 1).toString() + result;
			e -= 1;
		}
		// Display calculated result
		println(" Length : " + min);
		println(" Result : " + result);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Supersequence = Supersequence();
	// Test A
	// String a : abc
	// String b : fab
	// [fabc] Supersequence 
	// Result = 4
	task.findSCS("abc", "fab");
	// Test B
	// String a : project
	// String b : objects
	// [probjects]  
	// Result : 9
	task.findSCS("project", "objects");
	// Test C
	// String a : match
	// String b : attack
	// [mattachk,matatchk,mattackh] etc
	// Result : 8 (length of Supersequence)
	task.findSCS("match", "attack");
	// Test D
	// String a : abc
	// String b : abc
	// [abc] etc
	// Result : 3 
	task.findSCS("abc", "abc");
}

Output

 Given string a : abc
 Given string b : fab
 Length : 4
 Result : fabc
 Given string a : project
 Given string b : objects
 Length : 9
 Result : probjects
 Given string a : match
 Given string b : attack
 Length : 8
 Result : mattackh
 Given string a : abc
 Given string b : abc
 Length : 3
 Result : abc

Time Complexity

The time complexity of the provided algorithm is O(n * m), where n and m are the lengths of the input strings a and b. This is because the algorithm constructs a 2D dynamic programming table of size (n+1) x (m+1) and fills it based on certain conditions. The space complexity is also O(n * m) due to the storage of the dynamic programming table.

Last updated on by Kalkicode




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