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).