Given two strings X and Y, print the shortest string that has both X and Y as subsequences. If multiple shortest super-sequence 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
We have discussed how to print length of shortest possible super-sequence for two given strings here. In this post, we print the shortest super-sequence.
We have already discussed below algorithm to find length of shortest super-sequence 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 super-sequence 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 the implementation of above idea –
C++14
/* 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 Ystring 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 // 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--; } // 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--; } else { // Put current character of X in result str.push_back(X[i - 1]); // reduce values of i and index i--; } } // If Y reaches its end, put remaining characters // of X in the result string while (i > 0) { str.push_back(X[i - 1]); i--; } // If X reaches its end, put remaining characters // of Y in the result string while (j > 0) { str.push_back(Y[j - 1]); j--; } // reverse the string and return it reverse(str.begin(), str.end()); return str;}// Driver program to test above functionint main(){ string X = "AGGTAB"; string Y = "GXTXAYB"; cout << printShortestSuperSeq(X, Y); return 0;} |
Java
/* A dynamic programming based Java program print shortest supersequence of two strings */class GFG { // returns shortest supersequence of X and Y static 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[][] = new int[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.charAt(i - 1) == Y.charAt(j - 1)) { dp[i][j] = 1 + dp[i - 1][j - 1]; } else { dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]); } } } // Following code is used to print // shortest supersequence dp[m][n] s // tores the length of the shortest // supersequence of X and Y // 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.charAt(i - 1) == Y.charAt(j - 1)) { // Put current character in result str += (X.charAt(i - 1)); // reduce values of i, j and index i--; j--; } // 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 += (Y.charAt(j - 1)); // reduce values of j and index j--; } else { // Put current character of X in result str += (X.charAt(i - 1)); // reduce values of i and index i--; } } // If Y reaches its end, put remaining characters // of X in the result string while (i > 0) { str += (X.charAt(i - 1)); i--; } // If X reaches its end, put remaining characters // of Y in the result string while (j > 0) { str += (Y.charAt(j - 1)); j--; } // reverse the string and return it str = reverse(str); return str; } static String reverse(String input) { char[] temparray = input.toCharArray(); int left, right = 0; right = temparray.length - 1; for (left = 0; left < right; left++, right--) { // Swap values of left and right char temp = temparray[left]; temparray[left] = temparray[right]; temparray[right] = temp; } return String.valueOf(temparray); } // Driver code public static void main(String[] args) { String X = "AGGTAB"; String Y = "GXTXAYB"; System.out.println(printShortestSuperSeq(X, Y)); }} // This code is contributed by 29AjayKumar |
Python3
# A dynamic programming based Python3 program print# shortest supersequence of two strings# returns shortest supersequence of X and Ydef printShortestSuperSeq(m, n, x, y): # dp[i][j] contains length of shortest # supersequence for X[0..i-1] and Y[0..j-1] dp = [[0 for i in range(n + 1)] for j in range(m + 1)] # Fill table in bottom up manner for i in range(m + 1): for j in range(n + 1): # Below steps follow recurrence relation if i == 0: dp[i][j] = j elif j == 0: dp[i][j] = i elif 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 # string to store the shortest supersequence string = "" # Start from the bottom right corner and # add the characters to the output string i = m j = n while i * 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 string = x[i - 1] + string # reduce values of i, j and index i -= 1 j -= 1 # If current character in X and Y are different elif dp[i - 1][j] > dp[i][j - 1]: # Put current character of Y in result string = y[j - 1] + string # reduce values of j and index j -= 1 else: # Put current character of X in result string = x[i - 1] + string # reduce values of i and index i -= 1 # If Y reaches its end, put remaining characters # of X in the result string while i > 0: string = x[i - 1] + string i -= 1 # If X reaches its end, put remaining characters # of Y in the result string while j > 0: string = y[j - 1] + string j -= 1 return string# Driver Codeif __name__ == "__main__": x = "GXTXAYB" y = "AGGTAB" m = len(x) n = len(y) # Take the smaller string as x and larger one as y if m > n: x, y = y, x m, n = n, m print(*printShortestSuperSeq(m, n, x, y))# This code is contributed by# sanjeev2552 |
C#
/* A dynamic programming based C# program print shortest supersequence of two strings */using System;class GFG{ // returns shortest supersequence of X and Y static 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 = new int[m + 1, n + 1]; int i, j; // Fill table in bottom up manner for (i = 0; i <= m; i++) { for (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 + Math.Min(dp[i - 1, j], dp[i, j - 1]); } } } // Following code is used to print // shortest supersequence dp[m,n] s // tores the length of the shortest // supersequence of X and Y // string to store the shortest supersequence String str = ""; // Start from the bottom right corner and one by one // push characters in output string 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 += (X[i - 1]); // reduce values of i, j and index i--; j--; } // 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 += (Y[j - 1]); // reduce values of j and index j--; } else { // Put current character of X in result str += (X[i - 1]); // reduce values of i and index i--; } } // If Y reaches its end, put remaining characters // of X in the result string while (i > 0) { str += (X[i - 1]); i--; } // If X reaches its end, put remaining characters // of Y in the result string while (j > 0) { str += (Y[j - 1]); j--; } // reverse the string and return it str = reverse(str); return str; } static String reverse(String input) { char[] temparray = input.ToCharArray(); int left, right = 0; right = temparray.Length - 1; for (left = 0; left < right; left++, right--) { // Swap values of left and right char temp = temparray[left]; temparray[left] = temparray[right]; temparray[right] = temp; } return String.Join("",temparray); } // Driver code public static void Main(String[] args) { String X = "AGGTAB"; String Y = "GXTXAYB"; Console.WriteLine(printShortestSuperSeq(X, Y)); }}/* This code has been contributed by PrinciRaj1992*/ |
Javascript
<script>/* A dynamic programming based Javascript program print shortest supersequence of two strings */// returns shortest supersequence of X and Y function printShortestSuperSeq(X,Y){ let m = X.length; let n = Y.length; // dp[i][j] contains length of // shortest supersequence // for X[0..i-1] and Y[0..j-1] let dp = new Array(m + 1); for(let i=0;i<(m+1);i++) { dp[i]=new Array(n+1); for(let j=0;j<(n+1);j++) dp[i][j]=0; } // Fill table in bottom up manner for (let i = 0; i <= m; i++) { for (let 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 + Math.min(dp[i - 1][j], dp[i][j - 1]); } } } // Following code is used to print // shortest supersequence dp[m][n] s // tores the length of the shortest // supersequence of X and Y // string to store the shortest supersequence let str = ""; // Start from the bottom right corner and one by one // push characters in output string let 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 += (X[i-1]); // reduce values of i, j and index i--; j--; } // 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 += (Y[j-1]); // reduce values of j and index j--; } else { // Put current character of X in result str += (X[i-1]); // reduce values of i and index i--; } } // If Y reaches its end, put remaining characters // of X in the result string while (i > 0) { str += (X[i-1]); i--; } // If X reaches its end, put remaining characters // of Y in the result string while (j > 0) { str += (Y[j-1]); j--; } // reverse the string and return it str = reverse(str); return str;}function reverse(input){ let temparray = input.split(""); let left, right = 0; right = temparray.length - 1; for (left = 0; left < right; left++, right--) { // Swap values of left and right let temp = temparray[left]; temparray[left] = temparray[right]; temparray[right] = temp; } return (temparray).join("");}// Driver codelet X = "AGGTAB";let Y = "GXTXAYB";document.write(printShortestSuperSeq(X, Y));// This code is contributed by rag2127</script> |
Output
AGXGTXAYB
Time complexity of above solution is O(n2).
Auxiliary space used by the program is O(n2).
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