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)
{
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
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