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

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; ` `} ` |

*chevron_right*

*filter_none*

Output:

AGXGTXAYB

**Time complexity** of above solution is O(n^{2}).

**Auxiliary space** used by the program is O(n^{2}).

This article is contributed by **Aditya Goel**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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