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Code Dynamic Programming

# Shortest Common Supersequence

In computer science, the Shortest Common Supersequence (SCS) problem is a classic algorithmic problem. Given two strings, the goal is to find the shortest possible string that contains both input strings as subsequences.

## Problem Statement

For example, let's consider two strings: string a and string b. The SCS is defined as the shortest string s such that both a and b are subsequences of s. In other words, s contains all the characters of a and b, preserving their relative order, but it may also contain additional characters.

The goal is to find the length of the SCS, as well as the SCS itself. If multiple SCS strings exist, any one of them can be considered as the result.

## Algorithm

The algorithm for finding the length of the SCS can be implemented using dynamic programming. Let's consider two strings, a of length n and b of length m. We create a 2D array, dp, of size (n+1) x (m+1), where dp[i][j] represents the length of the SCS for the prefixes of a and b up to index i and j, respectively.

The algorithm follows these steps:

1. Initialize the array dp with appropriate values:
• dp[0][j] is set to j because when the length of a is zero, the SCS is obtained by adding all characters of b.
• dp[i][0] is set to i because when the length of b is zero, the SCS is obtained by adding all characters of a.
2. Iterate through each character of a and b:
• If the characters at position i-1 in a and j-1 in b are the same, set dp[i][j] to dp[i-1][j-1] + 1. This means that the current characters contribute one character to the SCS.
• Otherwise, set dp[i][j] to the minimum of dp[i][j-1] and dp[i-1][j] plus one. This accounts for the case where either a or b contributes a character to the SCS.
3. The final result is stored in dp[n][m], where n and m are the lengths of strings a and b, respectively.
Code solution
``````/*
Java program for
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 lengthSCS(String a, String b)
{
int n = a.length();
int m = b.length();
// 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);
// Display calculated result
System.out.println(" Result : " + dp[n][m]);
}
public static void main(String[] args)
{
Supersequence task = 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
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
C++ program for
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 lengthSCS(string a, string b)
{
int n = a.length();
int m = b.length();
// 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;
// Display calculated result
cout << " Result : " << dp[n][m] << endl;
}
};
int main()
{
Supersequence *task = 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
return 0;
}``````

#### Output

`````` Given string a : abc
Given string b : fab
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````// Include namespace system
using System;
/*
Csharp program for
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 lengthSCS(String a, String b)
{
int n = a.Length;
int m = b.Length;
// 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);
// Display calculated result
Console.WriteLine(" Result : " + dp[n,m]);
}
public static void Main(String[] args)
{
Supersequence task = 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
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````package main
import "fmt"
/*
Go program for
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 lengthSCS(a, b string) {
var n int = len(a)
var m int = len(b)
// 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)
// Display calculated result
fmt.Println(" Result : ", dp[n][m])
}
func main() {

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

#### Output

`````` Given string a : abc
Given string b : fab
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````<?php
/*
Php program for
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 lengthSCS(\$a, \$b)
{
\$n = strlen(\$a);
\$m = strlen(\$b);
// 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");
// Display calculated result
echo(" Result : ".\$dp[\$n][\$m].
"\n");
}
}

function main()
{
\$task = 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
}
main();``````

#### Output

`````` Given string a : abc
Given string b : fab
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````/*
Node JS program for
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
lengthSCS(a, b)
{
var n = a.length;
var m = b.length;
// 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);
// Display calculated result
console.log(" Result : " + dp[n][m]);
}
}

function main()
{
var task = 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
}
main();``````

#### Output

`````` Given string a : abc
Given string b : fab
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````#    Python 3 program for
#    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 lengthSCS(self, a, b) :
n = len(a)
m = len(b)
#  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)
#  Display calculated result
print(" Result : ", dp[n][m])

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
Result :  4
Given string a :  project
Given string b :  objects
Result :  9
Given string a :  match
Given string b :  attack
Result :  8
Given string a :  abc
Given string b :  abc
Result :  3``````
``````#    Ruby program for
#    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 lengthSCS(a, b)
n = a.length
m = b.length
#  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")
#  Display calculated result
print(" Result : ", dp[n][m], "\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
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3
``````
``````import scala.collection.mutable._;
/*
Scala program for
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 lengthSCS(a: String, b: String): Unit = {
var n: Int = a.length();
var m: Int = b.length();
// 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);
// Display calculated result
println(" Result : " + dp(n)(m));
}
}
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
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````
``````import Foundation;
/*
Swift 4 program for
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 lengthSCS(_ a1: String, _ b1: String)
{
let a = Array(a1);
let b = Array(b1);
let n: Int = a.count;
let m: Int = b.count;
// 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);
// Display calculated result
print(" Result : ", dp[n][m]);
}
}
func main()
{
let task: Supersequence = 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
}
main();``````

#### Output

`````` Given string a :  abc
Given string b :  fab
Result :  4
Given string a :  project
Given string b :  objects
Result :  9
Given string a :  match
Given string b :  attack
Result :  8
Given string a :  abc
Given string b :  abc
Result :  3``````
``````/*
Kotlin program for
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 lengthSCS(a: String, b: String): Unit
{
val n: Int = a.length;
val m: Int = b.length;
// Auxiliary space
val 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);
// Display calculated result
println(" Result : " + dp[n][m]);
}
}
fun main(args: Array < String > ): Unit
{
val task: Supersequence = 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
Result : 4
Given string a : project
Given string b : objects
Result : 9
Given string a : match
Given string b : attack
Result : 8
Given string a : abc
Given string b : abc
Result : 3``````

## Explanation

Let's analyze the code and understand how it solves the SCS problem. The code uses dynamic programming to build a table, dp, to store the lengths of SCS for different prefixes of strings a and b.

The algorithm initializes the table by setting the values of dp[0][j] and dp[i][0] to represent the length of the SCS when one of the strings is empty. Then, it iterates through the characters of a and b to fill in the rest of the table.

For each pair of characters at positions i-1 and j-1 in a and b, respectively, the algorithm checks if the characters are the same. If they are, it adds one to the length of the SCS for the prefixes a and b without considering these characters (dp[i-1][j-1] + 1).

If the characters are different, the algorithm considers two options: either include the character from a or include the character from b. It chooses the minimum length between these two options plus one (minValue(dp[i][j-1], dp[i-1][j]) + 1).

Finally, the result is stored in dp[n][m], which represents the length of the SCS for the entire strings a and b. The algorithm outputs the given strings and the calculated result.

## Time Complexity

The time complexity of this algorithm is O(n * m), where n and m are the lengths of strings a and b, respectively. This is because the algorithm iterates through all possible pairs of characters in a and b to fill in the dp table. Each iteration takes constant time operations, resulting in a total time complexity proportional to the product of the string lengths.

The space complexity of the algorithm is also O(n * m) since it uses a 2D array, dp, of size (n+1) x (m+1) to store the lengths of SCS for different prefixes of a and b.

The Shortest Common Supersequence problem is a well-known algorithmic problem that aims to find the shortest string containing two input strings as subsequences. The provided Java code demonstrates an efficient solution using dynamic programming. By building a table and considering different cases, the algorithm calculates the length of the SCS and outputs the result.

Understanding and implementing algorithms like the Shortest Common Supersequence can enhance your problem-solving skills and help you tackle similar challenges in the future.

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