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

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 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

// 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

# 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#

// 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);

    }
}
//

Javascript

<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++

// 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

Python3

# 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#

// 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

Javascript

<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. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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