Shortest possible combination of two strings

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

/* 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 indexs 
          i--; j--; index--; 
      } 
  
      // 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--;  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

// 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 indexs 
              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 

Output:

algorihythm

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 minimum of the adjacent positions. Finally, we backtrack the matrix to find the index vector traversing which would yield the shortest possible combination.

C++

// 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

// 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 

Output:

algorihythm

This article is contributed by Raghav Jajodia. 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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