# 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 ` `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 `

## Python3

 `# 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;  ` `            ``elif` `not` `j:  ` `                ``dp[i][j] ``=` `i;  ` `            ``elif` `(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 indexs  ` `            ``i ``-``=` `1` `            ``j ``-``=` `1` `            ``index ``-``=` `1` `         `  `        ``# If not same, then find the larger of two and  ` `        ``# go in the direction of larger value  ` `        ``elif` `(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#

 `// 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 indexs ` `              ``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); ` `        `  `    ``} ` `} ` `// This article is contributed by Ita_c. `

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 ` `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[], 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; ` `    ``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 index_a = ``new` `ArrayList<>(); ` `    ``static` `ArrayList 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 `

## C#

 `// 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 `

Output:

```algorihythm
```

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