# Printing Shortest Common Supersequence

Here given code implementation process.

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

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

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

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

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

#### 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``````

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