Compute the shortest string for a combination of two given strings such that the new string consist of both the strings as its subsequences.
Examples:
Input : a = "pear" b = "peach" Output : pearch pearch is the shorted string such that both pear and peach are its subsequences. Input : a = "geek" b = "code" Output : gecodek
We have discussed a solution to find length of the shortest supersequence in below post.
Shortest Common Supersequence
In this post, printing of supersequence is discussed. The solution is based on below recursive approach discussed in above post as an alternate method.
Let a[0..m-1] and b[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 a[] if (a[m-1] == b[n-1]) return 1 + SCS(a, b, 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(a, b, m-1, n), SCS(a, b, m, n-1) );
We build a DP array to store lengths. After building the DP array, we traverse from bottom right most position. The approach of printing is similar to printing LCS.
/* C++ program to print supersequence of two strings */
#include<bits/stdc++.h> using namespace std;
/* Prints super sequence of a[0..m-1] and b[0..n-1] */ void printSuperSeq(string &a, string &b)
{ int m = a.length(), n = b.length();
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 above recurrence
if (!i)
dp[i][j] = j;
else if (!j)
dp[i][j] = i;
else if (a[i-1] == b[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 supersequence
int index = dp[m][n];
// Create a string of size index+1 to store the result
string res(index+1, '\0' );
// Start from the right-most-bottom-most corner and
// one by one store characters in res[]
int i = m, j = n;
while (i > 0 && j > 0)
{
// If current character in a[] and b are same,
// then current character is part of LCS
if (a[i-1] == b[j-1])
{
// Put current character in result
res[index-1] = a[i-1];
// reduce values of i, j and index
i--; j--; index--;
}
// If not same, then find the smaller of two and
// go in the direction of smaller value
else if (dp[i-1][j] < dp[i][j-1])
{ res[index-1] = a[i-1]; i--; index--; }
else
{ res[index-1] = b[j-1]; j--; index--; }
}
// Copy remaining characters of string 'a'
while (i > 0)
{
res[index-1] = a[i-1]; i--; index--;
}
// Copy remaining characters of string 'b'
while (j > 0)
{
res[index-1] = b[j-1]; j--; index--;
}
// Print the result
cout << res;
} /* Driver program to test above function */ int main()
{ string a = "algorithm" , b = "rhythm" ;
printSuperSeq(a, b);
return 0;
} |
// Java program to print supersequence of two // strings public class GFG_1 {
String a , b;
// Prints super sequence of a[0..m-1] and b[0..n-1]
static void printSuperSeq(String a, String b)
{
int m = a.length(), n = b.length();
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 above recurrence
if (i == 0 )
dp[i][j] = j;
else if (j == 0 )
dp[i][j] = i;
else if (a.charAt(i- 1 ) == b.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 ]);
}
}
// Create a string of size index+1 to store the result
String res = "" ;
// Start from the right-most-bottom-most corner and
// one by one store characters in res[]
int i = m, j = n;
while (i > 0 && j > 0 )
{
// If current character in a[] and b are same,
// then current character is part of LCS
if (a.charAt(i- 1 ) == b.charAt(j- 1 ))
{
// Put current character in result
res = a.charAt(i- 1 ) + res;
// reduce values of i, j and indexes
i--;
j--;
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (dp[i- 1 ][j] < dp[i][j- 1 ])
{
res = a.charAt(i- 1 ) + res;
i--;
}
else
{
res = b.charAt(j- 1 ) + res;
j--;
}
}
// Copy remaining characters of string 'a'
while (i > 0 )
{
res = a.charAt(i- 1 ) + res;
i--;
}
// Copy remaining characters of string 'b'
while (j > 0 )
{
res = b.charAt(j- 1 ) + res;
j--;
}
// Print the result
System.out.println(res);
}
/* Driver program to test above function */
public static void main(String args[])
{
String a = "algorithm" ;
String b = "rhythm" ;
printSuperSeq(a, b);
}
} // This article is contributed by Sumit Ghosh |
# Python3 program to print supersequence of two strings # Prints super sequence of a[0..m-1] and b[0..n-1] def printSuperSeq(a, b):
m = len (a)
n = len (b)
dp = [[ 0 ] * (n + 1 ) for i in range (m + 1 )]
# Fill table in bottom up manner
for i in range ( 0 , m + 1 ):
for j in range ( 0 , n + 1 ):
# Below steps follow above recurrence
if not i:
dp[i][j] = j;
else if not j:
dp[i][j] = i;
else if (a[i - 1 ] = = b[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 supersequence
index = dp[m][n];
# Create a string of size index+1
# to store the result
res = [""] * (index)
# Start from the right-most-bottom-most corner
# and one by one store characters in res[]
i = m
j = n;
while (i > 0 and j > 0 ):
# If current character in a[] and b are same,
# then current character is part of LCS
if (a[i - 1 ] = = b[j - 1 ]):
# Put current character in result
res[index - 1 ] = a[i - 1 ];
# reduce values of i, j and indexes
i - = 1
j - = 1
index - = 1
# If not same, then find the larger of two and
# go in the direction of larger value
else if (dp[i - 1 ][j] < dp[i][j - 1 ]):
res[index - 1 ] = a[i - 1 ]
i - = 1
index - = 1
else :
res[index - 1 ] = b[j - 1 ]
j - = 1
index - = 1
# Copy remaining characters of string 'a'
while (i > 0 ):
res[index - 1 ] = a[i - 1 ]
i - = 1
index - = 1
# Copy remaining characters of string 'b'
while (j > 0 ):
res[index - 1 ] = b[j - 1 ]
j - = 1
index - = 1
# Print the result
print ("".join(res))
# Driver Code if __name__ = = '__main__' :
a = "algorithm"
b = "rhythm"
printSuperSeq(a, b)
# This code is contributed by ashutosh450 |
// C# program to print supersequence of two // strings using System;
public class GFG_1 {
// Prints super sequence of a[0..m-1] and b[0..n-1]
static void printSuperSeq( string a, string b)
{
int m = a.Length, n = b.Length;
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 above recurrence
if (i == 0)
dp[i,j] = j;
else if (j == 0 )
dp[i,j] = i;
else if (a[i-1] == b[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]);
}
}
// Create a string of size index+1 to store the result
string res = "" ;
// Start from the right-most-bottom-most corner and
// one by one store characters in res[]
int k = m, l = n;
while (k > 0 && l > 0)
{
// If current character in a[] and b are same,
// then current character is part of LCS
if (a[k-1] == b[l-1])
{
// Put current character in result
res = a[k-1] + res;
// reduce values of i, j and indexes
k--;
l--;
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (dp[k-1,l] < dp[k,l-1])
{
res = a[k-1] + res;
k--;
}
else
{
res = b[l-1] + res;
l--;
}
}
// Copy remaining characters of string 'a'
while (k > 0)
{
res = a[k-1] + res;
k--;
}
// Copy remaining characters of string 'b'
while (l > 0)
{
res = b[l-1] + res;
l--;
}
// Print the result
Console.WriteLine(res);
}
/* Driver program to test above function */
public static void Main()
{
string a = "algorithm" ;
string b = "rhythm" ;
printSuperSeq(a, b);
}
} // |
<script> // Javascript program to print supersequence of two // strings // Prints super sequence of a[0..m-1] and b[0..n-1]
function printSuperSeq(a,b)
{ let m = a.length, n = b.length;
let dp = new Array(m+1);
for (let i=0;i<m+1;i++)
dp[i]= new Array(n+1);
// Fill table in bottom up manner
for (let i = 0; i <= m; i++)
{
for (let j = 0; j <= n; j++)
{
// Below steps follow above recurrence
if (i == 0)
dp[i][j] = j;
else if (j == 0 )
dp[i][j] = i;
else if (a[i-1] == b[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]);
}
}
// Create a string of size index+1 to store the result
let res = "" ;
// Start from the right-most-bottom-most corner and
// one by one store characters in res[]
let i = m, j = n;
while (i > 0 && j > 0)
{
// If current character in a[] and b are same,
// then current character is part of LCS
if (a[i-1] == b[j-1])
{
// Put current character in result
res = a[i-1] + res;
// reduce values of i, j and indexes
i--;
j--;
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (dp[i-1][j] < dp[i][j-1])
{
res = a[i-1] + res;
i--;
}
else
{
res = b[j-1] + res;
j--;
}
}
// Copy remaining characters of string 'a'
while (i > 0)
{
res = a[i-1] + res;
i--;
}
// Copy remaining characters of string 'b'
while (j > 0)
{
res = b[j-1] + res;
j--;
}
// Print the result
document.write(res);
} /* Driver program to test above function */ let a = "algorithm" ;
let b = "rhythm" ;
printSuperSeq(a, b); // This code is contributed by ab2127 </script> |
Output:
algorihythm
Time Complexity: O(m*n)
Auxiliary Space: O(m*n), where m = length of first string , n = length of second string.
Solution based on LCS:
We build the 2D array using LCS solution. If the character at the two pointer positions is equal, we increment the length by 1, else we store the maximum of the adjacent positions. Finally, we backtrack the matrix to find the index vector traversing which would yield the shortest possible combination.
// C++ implementation to find shortest string for // a combination of two strings #include <bits/stdc++.h> using namespace std;
// Vector that store the index of string a and b vector< int > index_a;
vector< int > index_b;
// Subroutine to Backtrack the dp matrix to // find the index vector traversing which would // yield the shortest possible combination void index( int dp[][100], string a, string b,
int size_a, int size_b)
{ // Clear the index vectors
index_a.clear();
index_b.clear();
// Return if either of a or b is reduced
// to 0
if (size_a == 0 || size_b == 0)
return ;
// Push both to index_a and index_b with
// the respective a and b index
if (a[size_a - 1] == b[size_b - 1]) {
index(dp, a, b, size_a - 1, size_b - 1);
index_a.push_back(size_a - 1);
index_b.push_back(size_b - 1);
} else {
if (dp[size_a - 1][size_b] > dp[size_a]
[size_b - 1]) {
index(dp, a, b, size_a - 1, size_b);
} else {
index(dp, a, b, size_a, size_b - 1);
}
}
} // function to combine the strings to form // the shortest string void combine(string a, string b, int size_a,
int size_b)
{ int dp[100][100];
string ans = "" ;
int k = 0;
// Initialize the matrix to 0
memset (dp, 0, sizeof (dp));
// Store the increment of diagonally
// previous value if a[i-1] and b[j-1] are
// equal, else store the max of dp[i][j-1]
// and dp[i-1][j]
for ( int i = 1; i <= size_a; i++) {
for ( int j = 1; j <= size_b; j++) {
if (a[i - 1] == b[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = max(dp[i][j - 1],
dp[i - 1][j]);
}
}
}
// Get the Lowest Common Subsequence
int lcs = dp[size_a][size_b];
// Backtrack the dp array to get the index
// vectors of two strings, used to find
// the shortest possible combination.
index(dp, a, b, size_a, size_b);
int i, j = i = k;
// Build the string combination using the
// index found by backtracking
while (k < lcs) {
while (i < size_a && i < index_a[k]) {
ans += a[i++];
}
while (j < size_b && j < index_b[k]) {
ans += b[j++];
}
ans = ans + a[index_a[k]];
k++;
i++;
j++;
}
// Append the remaining characters in a
// to answer
while (i < size_a) {
ans += a[i++];
}
// Append the remaining characters in b
// to answer
while (j < size_b) {
ans += b[j++];
}
cout << ans;
} // Driver code int main()
{ string a = "algorithm" ;
string b = "rhythm" ;
// Store the length of string
int size_a = a.size();
int size_b = b.size();
combine(a, b, size_a, size_b);
return 0;
} |
// Java implementation to find shortest string for // a combination of two strings import java.util.ArrayList;
public class GFG_2 {
// Vector that store the index of string a and b
static ArrayList<Integer> index_a = new ArrayList<>();
static ArrayList<Integer> index_b = new ArrayList<>();
// Subroutine to Backtrack the dp matrix to
// find the index vector traversing which would
// yield the shortest possible combination
static void index( int dp[][], String a, String b,
int size_a, int size_b)
{
// Clear the index vectors
index_a.clear();
index_b.clear();
// Return if either of a or b is reduced
// to 0
if (size_a == 0 || size_b == 0 )
return ;
// Push both to index_a and index_b with
// the respective a and b index
if (a.charAt(size_a - 1 ) == b.charAt(size_b - 1 )) {
index(dp, a, b, size_a - 1 , size_b - 1 );
index_a.add(size_a - 1 );
index_b.add(size_b - 1 );
} else {
if (dp[size_a - 1 ][size_b] > dp[size_a]
[size_b - 1 ]) {
index(dp, a, b, size_a - 1 , size_b);
} else {
index(dp, a, b, size_a, size_b - 1 );
}
}
}
// function to combine the strings to form
// the shortest string
static void combine(String a, String b, int size_a,
int size_b)
{
int [][] dp = new int [ 100 ][ 100 ];
String ans = "" ;
int k = 0 ;
// Store the increment of diagonally
// previous value if a[i-1] and b[j-1] are
// equal, else store the max of dp[i][j-1]
// and dp[i-1][j]
for ( int i = 1 ; i <= size_a; i++) {
for ( int j = 1 ; j <= size_b; j++) {
if (a.charAt(i - 1 ) == b.charAt(j - 1 )) {
dp[i][j] = dp[i - 1 ][j - 1 ] + 1 ;
} else {
dp[i][j] = Math.max(dp[i][j - 1 ],
dp[i - 1 ][j]);
}
}
}
// Get the Lowest Common Subsequence
int lcs = dp[size_a][size_b];
// Backtrack the dp array to get the index
// vectors of two strings, used to find
// the shortest possible combination.
index(dp, a, b, size_a, size_b);
int i, j = i = k;
// Build the string combination using the
// index found by backtracking
while (k < lcs) {
while (i < size_a && i < index_a.get(k)) {
ans += a.charAt(i++);
}
while (j < size_b && j < index_b.get(k)) {
ans += b.charAt(j++);
}
ans = ans + a.charAt(index_a.get(k));
k++;
i++;
j++;
}
// Append the remaining characters in a
// to answer
while (i < size_a) {
ans += a.charAt(i++);
}
// Append the remaining characters in b
// to answer
while (j < size_b) {
ans += b.charAt(j++);
}
System.out.println(ans);
}
/* Driver program to test above function */
public static void main(String args[])
{
String a = "algorithm" ;
String b = "rhythm" ;
combine(a, b, a.length(),b.length());
}
} // This article is contributed by Sumit Ghosh |
# Python implementation to find shortest string for # a combination of two strings index_a = []
index_b = []
def index(dp, a, b, size_a, size_b):
if (size_a = = 0 or size_b = = 0 ):
return
if (a[size_a - 1 ] = = b[size_b - 1 ]):
index(dp, a, b, size_a - 1 , size_b - 1 )
index_a.append(size_a - 1 )
index_b.append(size_b - 1 )
else :
if (dp[size_a - 1 ][size_b] > dp[size_a][size_b - 1 ]):
index(dp, a, b, size_a - 1 , size_b)
else :
index(dp, a, b, size_a, size_b - 1 )
def combine(a, b, size_a, size_b):
dp = [[ 0 for i in range ( 100 )] for j in range ( 100 )]
ans = ""
k = 0
for i in range ( 1 , size_a + 1 ):
for j in range ( 1 , size_b + 1 ):
if (a[i - 1 ] = = b[j - 1 ]):
dp[i][j] = dp[i - 1 ][j - 1 ] + 1
else :
dp[i][j] = max (dp[i][j - 1 ], dp[i - 1 ][j])
lcs = dp[size_a][size_b]
index(dp, a, b, size_a, size_b)
j = i = k
while (k < lcs):
while (i < size_a and i < index_a[k]):
ans + = a[i];
i + = 1
while (j < size_b and j < index_b[k]):
ans + = b[j]
j + = 1
ans = ans + a[index_a[k]]
k + = 1
i + = 1
j + = 1
while (i < size_a):
ans + = a[i]
i + = 1
while (j < size_b):
ans + = b[j]
j + = 1
print (ans)
# Driver code a = "algorithm"
b = "rhythm"
size_a = len (a)
size_b = len (b)
combine(a, b, size_a, size_b) # This code is contributed by avanitrachhadiya2155 |
// C# implementation to find shortest string for // a combination of two strings using System;
using System.Collections.Generic;
class GFG
{ // Vector that store the index of string a and b
static List< int > index_a = new List< int >();
static List< int > index_b = new List< int >();
// Subroutine to Backtrack the dp matrix to
// find the index vector traversing which would
// yield the shortest possible combination
static void index( int [,]dp, String a, String b,
int size_a, int size_b)
{
// Clear the index vectors
index_a.Clear();
index_b.Clear();
// Return if either of a or b is reduced
// to 0
if (size_a == 0 || size_b == 0)
return ;
// Push both to index_a and index_b with
// the respective a and b index
if (a[size_a - 1] == b[size_b - 1])
{
index(dp, a, b, size_a - 1, size_b - 1);
index_a.Add(size_a - 1);
index_b.Add(size_b - 1);
}
else
{
if (dp[size_a - 1,size_b] > dp[size_a,
size_b - 1])
{
index(dp, a, b, size_a - 1, size_b);
}
else
{
index(dp, a, b, size_a, size_b - 1);
}
}
}
// function to combine the strings to form
// the shortest string
static void combine(String a, String b,
int size_a, int size_b)
{
int [,] dp = new int [100, 100];
String ans = "" ;
int k = 0, i, j;
// Store the increment of diagonally
// previous value if a[i-1] and b[j-1] are
// equal, else store the max of dp[i,j-1]
// and dp[i-1,j]
for (i = 1; i <= size_a; i++)
{
for (j = 1; j <= size_b; j++)
{
if (a[i-1] == b[j - 1])
{
dp[i, j] = dp[i - 1, j - 1] + 1;
}
else
{
dp[i, j] = Math.Max(dp[i, j - 1],
dp[i - 1, j]);
}
}
}
// Get the Lowest Common Subsequence
int lcs = dp[size_a, size_b];
// Backtrack the dp array to get the index
// vectors of two strings, used to find
// the shortest possible combination.
index(dp, a, b, size_a, size_b);
i = j = k;
// Build the string combination using the
// index found by backtracking
while (k < lcs)
{
while (i < size_a && i < index_a[k])
{
ans += a[i++];
}
while (j < size_b && j < index_b[k])
{
ans += b[j++];
}
ans = ans + a[index_a[k]];
k++;
i++;
j++;
}
// Append the remaining characters in a
// to answer
while (i < size_a)
{
ans += a[i++];
}
// Append the remaining characters in b
// to answer
while (j < size_b)
{
ans += b[j++];
}
Console.WriteLine(ans);
}
// Driver Code
public static void Main(String []args)
{
String a = "algorithm" ;
String b = "rhythm" ;
combine(a, b, a.Length,b.Length);
}
} // This code is contributed by Princi Singh |
<script> // JavaScript implementation to find shortest string for // a combination of two strings // Vector that store the index of string a and b let index_a =[]; let index_b = []; // Subroutine to Backtrack the dp matrix to // find the index vector traversing which would // yield the shortest possible combination function index(dp,a,b,size_a,size_b)
{ // Clear the index vectors
index_a=[];
index_b=[];
// Return if either of a or b is reduced
// to 0
if (size_a == 0 || size_b == 0)
return ;
// Push both to index_a and index_b with
// the respective a and b index
if (a[size_a - 1] == b[size_b - 1]) {
index(dp, a, b, size_a - 1, size_b - 1);
index_a.push(size_a - 1);
index_b.push(size_b - 1);
} else {
if (dp[size_a - 1][size_b] > dp[size_a]
[size_b - 1]) {
index(dp, a, b, size_a - 1, size_b);
} else {
index(dp, a, b, size_a, size_b - 1);
}
}
} // function to combine the strings to form // the shortest string function combine(a,b,size_a,size_b)
{ let dp = new Array(100);
for (let i=0;i<100;i++)
{
dp[i]= new Array(100);
for (let j=0;j<100;j++)
{
dp[i][j]=0;
}
}
let ans = "" ;
let k = 0;
// Store the increment of diagonally
// previous value if a[i-1] and b[j-1] are
// equal, else store the max of dp[i][j-1]
// and dp[i-1][j]
for (let i = 1; i <= size_a; i++) {
for (let j = 1; j <= size_b; j++) {
if (a[i - 1] == b[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i][j - 1],
dp[i - 1][j]);
}
}
}
// Get the Lowest Common Subsequence
let lcs = dp[size_a][size_b];
// Backtrack the dp array to get the index
// vectors of two strings, used to find
// the shortest possible combination.
index(dp, a, b, size_a, size_b);
let i, j = i = k;
// Build the string combination using the
// index found by backtracking
while (k < lcs) {
while (i < size_a && i < index_a[k]) {
ans += a[i++];
}
while (j < size_b && j < index_b[k]) {
ans += b[j++];
}
ans = ans + a[index_a[k]];
k++;
i++;
j++;
}
// Append the remaining characters in a
// to answer
while (i < size_a) {
ans += a[i++];
}
// Append the remaining characters in b
// to answer
while (j < size_b) {
ans += b[j++];
}
document.write(ans+ "<br>" );
} /* Driver program to test above function */ let a = "algorithm" ;
let b = "rhythm" ;
combine(a, b, a.length,b.length); // This code is contributed by patel2127 </script> |
Output:
algorihythm
Time Complexity : O(n2)
Auxiliary Space: O(max(size_a,size_b)) //since dp array size is 101*101 which is constant
This article is contributed by Raghav Jajodia.