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

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:

**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)