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Lexicographically largest possible by merging two strings by adding one character at a time

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Given two strings S and T, the task is to merge these two strings by adding one character at a time from the beginning of either string to form a resultant string. The resultant string should be lexicographically the largest string that can be formed by merging the strings S and T

Examples:

Input: S = “dbcbb”, T = “cdbbb”
Output : dcdbcbbbbb

Input : S = geeks“, T = forgeeks”
Output : gforgeekseeks

Approach: The simplest idea to solve the problem is to greedily select the first characters from the string which is lexicographically bigger than others. Therefore, Greedy algorithm and Recursion can be used to solve the problem. Follow the steps below to solve the problem:

  •  If either of the string lengths is 0, then return S + T as the answer.
  • If S is lexicographically larger than T, then return S[0] + largestMerge(S.substr(1), T).
  • Otherwise, take the first character of T and call the recursive function largestMerge(S, T.substr(1)).

Below is the implementation of the above approach:

C++




// C++ program for the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Recursive  bfunction for finding
// the lexicographically largest string
string largestMerge(string s1, string s2)
{
    // If either of the string length is 0,
    // return the other string
    if (s1.size() == 0 || s2.size() == 0)
        return s1 + s2;
 
    // If s1 is lexicographically
    // larger than s2
    if (s1 > s2) {
 
        // Take first character of s1
        // and call the function
        return s1[0] + largestMerge(s1.substr(1), s2);
    }
 
    // Take first character of s2
    // and recursively call function for
    // remaining string
    return s2[0] + largestMerge(s1, s2.substr(1));
}
 
// Driver Code
int main()
{
    // Given Input
    string s1 = "geeks";
    string s2 = "forgeeks";
 
    // Function Call
    cout << largestMerge(s1, s2) << endl;
 
    return 0;
}


Java




// Java program for the approach
public class Main
{
    // Recursive  bfunction for finding
    // the lexicographically largest string
    static String largestMerge(String s1, String s2)
    {
       
        // If either of the string length is 0,
        // return the other string
        if (s1.length() == 0 || s2.length() == 0)
            return s1 + s2;
  
        // If s1 is lexicographically
        // larger than s2
        if (s1.compareTo(s2) > 0) {
  
            // Take first character of s1
            // and call the function
            return s1.charAt(0)
                + largestMerge(s1.substring(1), s2);
        }
  
        // Take first character of s2
        // and recursively call function for
        // remaining string
        return s2.charAt(0) + largestMerge(s1, s2.substring(1));
    }
     
    public static void main(String[] args)
    {
       
        // Given Input
        String s1 = "geeks";
        String s2 = "forgeeks";
  
        // Function Call
        System.out.print(largestMerge(s1, s2));
    }
}
 
// This code is contributed by divyesh072019.


Python3




# Python program for the above approach
 
# Recursive function for finding
# the lexicographically largest string
def largestMerge(s1, s2):
 
    # If either of the string length is 0,
    # return the other string
    if len(s1) == 0 or len(s2) == 0:
        return s1+s2
 
    # If s1 is lexicographically
    # larger than s2
    if(s1 > s2):
       
        # Take first character of s1
        # and call the function
        return s1[0]+largestMerge(s1[1:], s2)
 
    # Take first character of s2
    # and recursively call function for
    # remaining string
    return s2[0]+largestMerge(s1, s2[1:])
 
 
# Driver code
if __name__ == '__main__':
   
    # Given Input
    s1 = "geeks"
    s2 = "forgeeks"
 
    # Function call
    print(largestMerge(s1, s2))
     
# This code is contributed by MuskanKalra1


C#




// C# program for the approach
using System;
class GFG {
    // Recursive  bfunction for finding
    // the lexicographically largest string
    static string largestMerge(string s1, string s2)
    {
        // If either of the string length is 0,
        // return the other string
        if (s1.Length == 0 || s2.Length == 0)
            return s1 + s2;
 
        // If s1 is lexicographically
        // larger than s2
        if (string.Compare(s1, s2) == 1) {
 
            // Take first character of s1
            // and call the function
            return s1[0]
                + largestMerge(s1.Substring(1), s2);
        }
 
        // Take first character of s2
        // and recursively call function for
        // remaining string
        return s2[0] + largestMerge(s1, s2.Substring(1));
    }
 
    // Driver Code
    public static void Main()
    {
        // Given Input
        string s1 = "geeks";
        string s2 = "forgeeks";
 
        // Function Call
        Console.Write(largestMerge(s1, s2));
    }
}
 
// This code is contributed by ukasp.


Javascript




<script>
    // JavaScript program for the approach
 
    // Recursive bfunction for finding
    // the lexicographically largest string
    const largestMerge = (s1, s2) =>
    {
     
        // If either of the string length is 0,
        // return the other string
        if (s1.length == 0 || s2.length == 0)
            return s1 + s2;
 
        // If s1 is lexicographically
        // larger than s2
        if (s1 > s2) {
 
            // Take first character of s1
            // and call the function
            return s1[0] + largestMerge(s1.substr(1), s2);
        }
 
        // Take first character of s2
        // and recursively call function for
        // remaining string
        return s2[0] + largestMerge(s1, s2.substr(1));
    }
 
    // Driver Code
    // Given Input
    s1 = "geeks";
    s2 = "forgeeks";
 
    // Function Call
    document.write(largestMerge(s1, s2));
     
    // This code is contributed by rakeshsahni
</script>


Output

gforgeekseeks

Time Complexity: O(M×N), where M and N are the length of string s1 and s2 respectively.
Auxiliary Space: O(1)



Last Updated : 05 Oct, 2021
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