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# Shortest Common Supersequence

Given two strings str1 and str2, the task is to find the length of the shortest string that has both str1 and str2 as subsequences.

Examples :

Input:   str1 = "geek",  str2 = "eke"
Output: 5
Explanation:
String "geeke" has both string "geek"
and "eke" as subsequences.

Input:   str1 = "AGGTAB",  str2 = "GXTXAYB"
Output:  9
Explanation:
String "AGXGTXAYB" has both string
"AGGTAB" and "GXTXAYB" as subsequences.
Recommended Practice

Method 1: This problem is closely related to longest common subsequence problem.
Below are steps.

1. Find Longest Common Subsequence (lcs) of two given strings. For example, lcs of “geek” and “eke” is “ek”.
2. Insert non-lcs characters (in their original order in strings) to the lcs found above, and return the result. So “ek” becomes “geeke” which is shortest common supersequence.
Let us consider another example, str1 = “AGGTAB” and str2 = “GXTXAYB”. LCS of str1 and str2 is “GTAB”. Once we find LCS, we insert characters of both strings in order and we get “AGXGTXAYB”

How does this work?

We need to find a string that has both strings as subsequences and is the shortest such string. If both strings have all characters different, then result is sum of lengths of two given strings. If there are common characters, then we don’t want them multiple times as the task is to minimize length. Therefore, we first find the longest common subsequence, take one occurrence of this subsequence and add extra characters.

Length of the shortest supersequence
= (Sum of lengths of given two strings)
- (Length of LCS of two given strings)

Below is the implementation of above idea. The below implementation only finds length of the shortest super sequence.

## C++

 // C++ program to find length of the // shortest supersequence #include using namespace std;   // Utility function to get max // of 2 integers int max(int a, int b) { return (a > b) ? a : b; }   // Returns length of LCS for // X[0..m - 1], Y[0..n - 1] int lcs(char* X, char* Y, int m, int n);   // Function to find length of the // shortest supersequence of X and Y. int shortestSuperSequence(char* X, char* Y) {     int m = strlen(X), n = strlen(Y);       // find lcs     int l = lcs(X, Y, m, n);       // Result is sum of input string     // lengths - length of lcs     return (m + n - l); }   // Returns length of LCS // for X[0..m - 1], Y[0..n - 1] int lcs(char* X, char* Y, int m, int n) {     int L[m + 1][n + 1];     int i, j;       // Following steps build L[m + 1][n + 1]     // in bottom up fashion. Note that     // L[i][j] contains length of LCS of     // X[0..i - 1] and Y[0..j - 1]     for (i = 0; i <= m; i++) {         for (j = 0; j <= n; j++) {             if (i == 0 || j == 0)                 L[i][j] = 0;               else if (X[i - 1] == Y[j - 1])                 L[i][j] = L[i - 1][j - 1] + 1;               else                 L[i][j] = max(L[i - 1][j], L[i][j - 1]);         }     }       // L[m][n] contains length of LCS     // for X[0..n - 1] and Y[0..m - 1]     return L[m][n]; }   // Driver code int main() {     char X[] = "AGGTAB";     char Y[] = "GXTXAYB";       cout << "Length of the shortest supersequence is "          << shortestSuperSequence(X, Y) << endl;       return 0; }   // This code is contributed by Akanksha Rai

## C

 // C program to find length of // the shortest supersequence #include #include   // Utility function to get // max of 2 integers int max(int a, int b) { return (a > b) ? a : b; }   // Returns length of LCS for // X[0..m - 1], Y[0..n - 1] int lcs(char* X, char* Y, int m, int n);   // Function to find length of the // shortest supersequence of X and Y. int shortestSuperSequence(char* X, char* Y) {     int m = strlen(X), n = strlen(Y);       // find lcs     int l = lcs(X, Y, m, n);       // Result is sum of input string     // lengths - length of lcs     return (m + n - l); }   // Returns length of LCS // for X[0..m - 1], Y[0..n - 1] int lcs(char* X, char* Y, int m, int n) {     int L[m + 1][n + 1];     int i, j;       // Following steps build L[m + 1][n + 1]     // in bottom up fashion. Note that     // L[i][j] contains length of LCS of     // X[0..i - 1] and Y[0..j - 1]     for (i = 0; i <= m; i++) {         for (j = 0; j <= n; j++) {             if (i == 0 || j == 0)                 L[i][j] = 0;               else if (X[i - 1] == Y[j - 1])                 L[i][j] = L[i - 1][j - 1] + 1;               else                 L[i][j] = max(L[i - 1][j], L[i][j - 1]);         }     }       // L[m][n] contains length of LCS     // for X[0..n - 1] and Y[0..m - 1]     return L[m][n]; }   // Driver code int main() {     char X[] = "AGGTAB";     char Y[] = "GXTXAYB";       printf("Length of the shortest supersequence is %d\n",            shortestSuperSequence(X, Y));       return 0; }

## Java

 // Java program to find length of // the shortest supersequence import java.io.*; class GFG {       // Function to find length of the     // shortest supersequence of X and Y.     static int shortestSuperSequence(String X, String Y)     {         int m = X.length();         int n = Y.length();           // find lcs         int l = lcs(X, Y, m, n);           // Result is sum of input string         // lengths - length of lcs         return (m + n - l);     }       // Returns length of LCS     // for X[0..m - 1], Y[0..n - 1]     static int lcs(String X, String Y, int m, int n)     {         int[][] L = new int[m + 1][n + 1];         int i, j;           // Following steps build L[m + 1][n + 1]         // in bottom up fashion. Note that         // L[i][j] contains length of LCS         // of X[0..i - 1]and Y[0..j - 1]         for (i = 0; i <= m; i++) {             for (j = 0; j <= n; j++) {                 if (i == 0 || j == 0)                     L[i][j] = 0;                   else if (X.charAt(i - 1) == Y.charAt(j - 1))                     L[i][j] = L[i - 1][j - 1] + 1;                   else                     L[i][j] = Math.max(L[i - 1][j],                                        L[i][j - 1]);             }         }           // L[m][n] contains length of LCS         // for X[0..n - 1] and Y[0..m - 1]         return L[m][n];     }       // Driver code     public static void main(String args[])     {         String X = "AGGTAB";         String Y = "GXTXAYB";           System.out.println("Length of the shortest "                            + "supersequence is "                            + shortestSuperSequence(X, Y));     } }   // This article is contributed by Sumit Ghosh

## Python3

 # Python program to find length # of the shortest supersequence   # Function to find length of the # shortest supersequence of X and Y.     def shortestSuperSequence(X, Y):     m = len(X)     n = len(Y)     l = lcs(X, Y, m, n)       # Result is sum of input string     # lengths - length of lcs     return (m + n - l)   # Returns length of LCS for # X[0..m - 1], Y[0..n - 1]     def lcs(X, Y, m, n):     L = [[0] * (n + 2) for i in          range(m + 2)]       # Following steps build L[m + 1][n + 1]     # in bottom up fashion. Note that L[i][j]     # contains length of LCS of X[0..i - 1]     # and Y[0..j - 1]     for i in range(m + 1):           for j in range(n + 1):               if (i == 0 or j == 0):                 L[i][j] = 0               elif (X[i - 1] == Y[j - 1]):                 L[i][j] = L[i - 1][j - 1] + 1               else:                 L[i][j] = max(L[i - 1][j],                               L[i][j - 1])       # L[m][n] contains length of     # LCS for X[0..n - 1] and Y[0..m - 1]     return L[m][n]     # Driver code X = "AGGTAB" Y = "GXTXAYB"   print("Length of the shortest supersequence is %d"       % shortestSuperSequence(X, Y))   # This code is contributed by Ansu Kumari

## C#

 // C# program to find length of // the shortest supersequence using System;   class GFG {     // Function to find length of the     // shortest supersequence of X and Y.     static int shortestSuperSequence(String X, String Y)     {         int m = X.Length;         int n = Y.Length;           // find lcs         int l = lcs(X, Y, m, n);           // Result is sum of input string         // lengths - length of lcs         return (m + n - l);     }       // Returns length of LCS for     // X[0..m - 1], Y[0..n - 1]     static int lcs(String X, String Y, int m, int n)     {         int[, ] L = new int[m + 1, n + 1];         int i, j;           // Following steps build L[m + 1][n + 1]         // in bottom up fashion.Note that         // L[i][j] contains length of LCS of         // X[0..i - 1] and Y[0..j - 1]         for (i = 0; i <= m; i++) {             for (j = 0; j <= n; j++) {                 if (i == 0 || j == 0)                     L[i, j] = 0;                   else if (X[i - 1] == Y[j - 1])                     L[i, j] = L[i - 1, j - 1] + 1;                   else                     L[i, j] = Math.Max(L[i - 1, j],                                        L[i, j - 1]);             }         }           // L[m][n] contains length of LCS         // for X[0..n - 1] and Y[0..m - 1]         return L[m, n];     }       // Driver code     public static void Main()     {         String X = "AGGTAB";         String Y = "GXTXAYB";           Console.WriteLine("Length of the shortest"                           + "supersequence is "                           + shortestSuperSequence(X, Y));     } }   // This code is contributed by Sam007



## Javascript



Output

Length of the shortest supersequence is 9

Time Complexity: O(m*n).
Auxiliary Space: O(m*n)

Method 2: A simple analysis yields below simple recursive solution.

Let X[0..m - 1] and Y[0..n - 1] be two
strings and m and n 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) );

Below is simple naive recursive solution based on above recursive formula.

## C++

 // A Naive recursive C++ program to find // length of the shortest supersequence #include using namespace std;   int superSeq(char* X, char* Y, int m, int n) {     if (!m)         return n;     if (!n)         return m;       if (X[m - 1] == Y[n - 1])         return 1 + superSeq(X, Y, m - 1, n - 1);       return 1            + min(superSeq(X, Y, m - 1, n),                  superSeq(X, Y, m, n - 1)); }   // Driver Code int main() {     char X[] = "AGGTAB";     char Y[] = "GXTXAYB";     cout << "Length of the shortest supersequence is "          << superSeq(X, Y, strlen(X), strlen(Y));     return 0; }

## Java

 // A Naive recursive Java program to find // length of the shortest supersequence import java.io.*; class GFG {     static int superSeq(String X, String Y, int m, int n)     {         if (m == 0)             return n;         if (n == 0)             return m;           if (X.charAt(m - 1) == Y.charAt(n - 1))             return 1 + superSeq(X, Y, m - 1, n - 1);           return 1             + Math.min(superSeq(X, Y, m - 1, n),                        superSeq(X, Y, m, n - 1));     }       // Driver code     public static void main(String args[])     {         String X = "AGGTAB";         String Y = "GXTXAYB";         System.out.println(             "Length of the shortest"             + "supersequence is: "             + superSeq(X, Y, X.length(), Y.length()));     } }   // This article is contributed by Sumit Ghosh

## Python3

 # A Naive recursive python program to find # length of the shortest supersequence     def superSeq(X, Y, m, n):     if (not m):         return n     if (not n):         return m       if (X[m - 1] == Y[n - 1]):         return 1 + superSeq(X, Y, m - 1, n - 1)       return 1 + min(superSeq(X, Y, m - 1, n),                    superSeq(X, Y, m, n - 1))     # Driver Code X = "AGGTAB" Y = "GXTXAYB" print("Length of the shortest supersequence is %d"       % superSeq(X, Y, len(X), len(Y)))   # This code is contributed by Ansu Kumari

## C#

 // A Naive recursive C# program to find // length of the shortest supersequence using System;   class GFG {     static int superSeq(String X, String Y, int m, int n)     {         if (m == 0)             return n;         if (n == 0)             return m;           if (X[m - 1] == Y[n - 1])             return 1 + superSeq(X, Y, m - 1, n - 1);           return 1             + Math.Min(superSeq(X, Y, m - 1, n),                        superSeq(X, Y, m, n - 1));     }       // Driver Code     public static void Main()     {         String X = "AGGTAB";         String Y = "GXTXAYB";         Console.WriteLine(             "Length of the shortest supersequence is: "             + superSeq(X, Y, X.Length, Y.Length));     } }   // This code is contributed by Sam007



## Javascript



Output

Length of the shortest supersequence is 9

Time Complexity: O(2min(m, n)). Since there are overlapping subproblems
Auxiliary Space: O(min(m, n)), due to recursive call stack

Method 3: We can efficiently solve this recursive problem using Dynamic Programming.
Below is the Dynamic Programming-based implementation.

## C++

 // A dynamic programming based C++ program to // find length of the shortest supersequence #include using namespace std;   // Returns length of the shortest // supersequence of X and Y int superSeq(char* X, char* Y, int m, int n) {     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 (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]);         }     }       return dp[m][n]; }   // Driver Code int main() {     char X[] = "AGGTAB";     char Y[] = "GXTXAYB";     cout << "Length of the shortest supersequence is "          << superSeq(X, Y, strlen(X), strlen(Y));     return 0; }

## Java

 // A dynamic programming based Java program to // find length of the shortest supersequence import java.io.*; class GFG {       // Returns length of the shortest     // supersequence of X and Y     static int superSeq(String X, String Y, int m, int n)     {         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 (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]);             }         }           return dp[m][n];     }       // Driver Code     public static void main(String args[])     {         String X = "AGGTAB";         String Y = "GXTXAYB";         System.out.println(             "Length of the shortest supersequence is "             + superSeq(X, Y, X.length(), Y.length()));     } }   // This article is contributed by Sumit Ghosh

## Python3

 # A dynamic programming based python program # to find length of the shortest supersequence   # Returns length of the shortest supersequence of X and Y     def superSeq(X, Y, m, n):     dp = [[0] * (n + 2) for i in range(m + 2)]       # Fill table in bottom up manner     for i in range(m + 1):         for j in range(n + 1):               # Below steps follow above recurrence             if (not i):                 dp[i][j] = j             elif (not j):                 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])       return dp[m][n]     # Driver Code X = "AGGTAB" Y = "GXTXAYB" print("Length of the shortest supersequence is %d"       % superSeq(X, Y, len(X), len(Y)))   # This code is contributed by Ansu Kumari

## C#

 // A dynamic programming based C# program to // find length of the shortest supersequence using System;   class GFG {     // Returns length of the shortest     // supersequence of X and Y     static int superSeq(String X, String Y, int m, int n)     {         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 (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]);             }         }           return dp[m, n];     }       // Driver code     public static void Main()     {         String X = "AGGTAB";         String Y = "GXTXAYB";         Console.WriteLine(             "Length of the shortest supersequence is "             + superSeq(X, Y, X.Length, Y.Length));     } }   // This code is contributed by Sam007



## Javascript



Output

Length of the shortest supersequence is 9

Time Complexity: O(m*n).
Auxiliary Space: O(m*n)
Thanks to Gaurav Ahirwar for suggesting this solution.

Method 4 (Top Down Memoization Approach):
The idea is to follow the simple recursive solution, and use a lookup table to avoid re-computations. Before computing the result for input, we check if the result is already computed or not. If already computed, we return that result.

## C++

 // A dynamic programming based C++ program // to find length of the shortest supersequence   // Returns length of the // shortest supersequence of X and Y #include using namespace std;   int superSeq(string X, string Y, int n, int m,              vector > lookup) {       if (m == 0 || n == 0) {         lookup[n][m] = n + m;     }       if (lookup[n][m] == 0)         if (X[n - 1] == Y[m - 1]) {             lookup[n][m]                 = superSeq(X, Y, n - 1, m - 1, lookup) + 1;         }           else {             lookup[n][m]                 = min(superSeq(X, Y, n - 1, m, lookup) + 1,                       superSeq(X, Y, n, m - 1, lookup) + 1);         }       return lookup[n][m]; }   // Driver Code int main() {     string X = "AGGTB";     string Y = "GXTXAYB";       vector > lookup(         X.size() + 1, vector(Y.size() + 1, 0));       cout << "Length of the shortest supersequence is "          << superSeq(X, Y, X.size(), Y.size(), lookup)          << endl;       return 0; }       // This code is contributed by niraj gusain

## Java

 // A dynamic programming based JAVA program // to find length of the shortest supersequence   // Returns length of the // shortest supersequence of X and Y import java.util.*;   class GFG {       static int superSeq(String X, String Y, int n, int m,                         int[][] lookup)     {           if (m == 0 || n == 0) {             lookup[n][m] = n + m;         }           if (lookup[n][m] == 0)             if (X.charAt(n - 1) == Y.charAt(m - 1)) {                 lookup[n][m]                     = superSeq(X, Y, n - 1, m - 1, lookup)                       + 1;             }               else {                 lookup[n][m] = Math.min(                     superSeq(X, Y, n - 1, m, lookup) + 1,                     superSeq(X, Y, n, m - 1, lookup) + 1);             }           return lookup[n][m];     }       // Driver Code     public static void main(String[] args)     {         String X = "AGGTB";         String Y = "GXTXAYB";           int[][] lookup             = new int[X.length() + 1][Y.length() + 1];           System.out.print(             "Length of the shortest supersequence is "             + superSeq(X, Y, X.length(), Y.length(), lookup)             + "\n");     } }   // This code contributed by umadevi9616

## Python3

 # A dynamic programming based python program # to find length of the shortest supersequence   # Returns length of the # shortest supersequence of X and Y   def superSeq(X,Y,n,m,lookup):           if m==0 or n==0:         lookup[n][m] = n+m       if (lookup[n][m] == 0):             if X[n-1]==Y[m-1]:             lookup[n][m] = superSeq(X,Y,n-1,m-1,lookup)+1               else:             lookup[n][m] = min(superSeq(X,Y,n-1,m,lookup)+1,                                superSeq(X,Y,n,m-1,lookup)+1)           return lookup[n][m]           # Driver Code X = "AGGTAB" Y = "GXTXAYB"   lookup = [[0 for j in range(len(Y)+1)]for i in range(len(X)+1)] print("Length of the shortest supersequence is {}"       .format(superSeq(X,Y,len(X),len(Y),lookup)))   # This code is contributed by Tanmay Ambadkar

## C#

 // A dynamic programming based C# program // to find length of the shortest supersequence   // Returns length of the // shortest supersequence of X and Y using System;   public class GFG {       static int superSeq(String X, String Y, int n,                         int m, int[,] lookup)     {           if (m == 0 || n == 0) {             lookup[n, m] = n + m;         }           if (lookup[n, m] == 0)             if (X[n - 1] == Y[m - 1]) {                 lookup[n, m] = superSeq(X, Y, n - 1,                                        m - 1, lookup) + 1;             }               else {                 lookup[n, m] = Math.Min(superSeq(X, Y, n - 1, m, lookup) +                                        1, superSeq(X, Y, n, m - 1, lookup) +                                        1);             }           return lookup[n, m];     }       // Driver Code     public static void Main(String[] args) {         String X = "AGGTB";         String Y = "GXTXAYB";           int[,] lookup = new int[X.Length + 1,Y.Length + 1];           Console.Write(                 "Length of the shortest supersequence is " +           superSeq(X, Y, X.Length, Y.Length, lookup) + "\n");     } }   // This code is contributed by gauravrajput1

## Javascript



Output

Length of the shortest supersequence is 9

Time Complexity: O(n*m)
Auxiliary Space: O(n*m)

Exercise:
Extend the above program to print shortest super sequence also using function to print LCS
Please refer Printing Shortest Common Supersequence for solution
References:
https://en.wikipedia.org/wiki/Shortest_common_supersequence