# Printing Shortest Common Supersequence

Given two strings X and Y, print the shortest string that has both X and Y as subsequences. If multiple shortest supersequence exists, print any one of them.

Examples:

```Input: X = "AGGTAB",  Y = "GXTXAYB"
Output: "AGXGTXAYB" OR "AGGXTXAYB"
OR Any string that represents shortest
supersequence of X and Y

Input: X = "HELLO",  Y = "GEEK"
Output: "GEHEKLLO" OR "GHEEKLLO"
OR Any string that represents shortest
supersequence of X and Y
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed how to print length of shortest possible supersequence for two given strings here. In this post, we print the shortest supersequence.

We have already discussed below algorithm to find length of shortest supersequence in previous post-

```Let X[0..m-1] and Y[0..n-1] be two strings and m and be respective
lengths.

if (m == 0) return n;
if (n == 0) return m;

// If last characters are same, then add 1 to result and
// recur for X[]
if (X[m-1] == Y[n-1])
return 1 + SCS(X, Y, m-1, n-1);

// Else find shortest of following two
//  a) Remove last character from X and recur
//  b) Remove last character from Y and recur
else return 1 + min( SCS(X, Y, m-1, n), SCS(X, Y, m, n-1) );
```

The following table shows steps followed by the above algorithm if we solve it in bottom-up manner using Dynamic Programming for strings X = “AGGTAB” and Y = “GXTXAYB”,

Using the DP solution matrix, we can easily print shortest supersequence of two strings by following below steps –

```We start from the bottom-right most cell of the matrix and
push characters in output string based on below rules-

1. If the characters corresponding to current cell (i, j)
in X and Y are same, then the character is part of shortest
supersequence. We append it in output string and move
diagonally to next cell (i.e. (i - 1, j - 1)).

2. If the characters corresponding to current cell (i, j)
in X and Y are different, we have two choices -

If matrix[i - 1][j] > matrix[i][j - 1],
we add character corresponding to current
cell (i, j) in string Y in output string
and move to the left cell i.e. (i, j - 1)
else
we add character corresponding to current
cell (i, j) in string X in output string
and move to the top cell i.e. (i - 1, j)

3. If string Y reaches its end i.e. j = 0, we add remaining
characters of string X in the output string
else if string X reaches its end i.e. i = 0, we add
remaining characters of string Y in the output string.
```

Below is C++ implementation of above idea –

```/* A dynamic programming based C++ program print
shortest supersequence of two strings */
#include <bits/stdc++.h>
using namespace std;

// returns shortest supersequence of X and Y
string printShortestSuperSeq(string X, string Y)
{
int m = X.length();
int n = Y.length();

// dp[i][j] contains length of shortest supersequence
// for X[0..i-1] and Y[0..j-1]
int dp[m + 1][n + 1];

// Fill table in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// Below steps follow recurrence relation
if(i == 0)
dp[i][j] = j;
else if(j == 0)
dp[i][j] = i;
else if(X[i - 1] == Y[j - 1])
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
}
}

// Following code is used to print shortest supersequence

// dp[m][n] stores the length of the shortest supersequence
// of X and Y
int index = dp[m][n];

// string to store the shortest supersequence
string str;

// Start from the bottom right corner and one by one
// push characters in output string
int i = m, j = n;
while (i > 0 && j > 0)
{
// If current character in X and Y are same, then
// current character is part of shortest supersequence
if (X[i - 1] == Y[j - 1])
{
// Put current character in result
str.push_back(X[i - 1]);

// reduce values of i, j and index
i--, j--, index--;
}

// If current character in X and Y are different
else if (dp[i - 1][j] > dp[i][j - 1])
{
// Put current character of Y in result
str.push_back(Y[j - 1]);

// reduce values of j and index
j--, index--;
}
else
{
// Put current character of X in result
str.push_back(X[i - 1]);

// reduce values of i and index
i--, index--;
}
}

// If Y reaches its end, put remaining characters
// of X in the result string
while (i > 0)
{
str.push_back(X[i - 1]);
i--, index--;
}

// If X reaches its end, put remaining characters
// of Y in the result string
while (j > 0)
{
str.push_back(Y[j - 1]);
j--, index--;
}

// reverse the string and return it
reverse(str.begin(), str.end());
return str;
}

// Driver program to test above function
int main()
{
string X = "AGGTAB";
string Y = "GXTXAYB";

cstr << printShortestSuperSeq(X, Y);

return 0;
}
```

Output:

```AGXGTXAYB
```

Time complexity of above solution is O(n2).
Auxiliary space used by the program is O(n2).

# GATE CS Corner    Company Wise Coding Practice

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