# Find the winner of the game to build the lexicographically smaller string

• Last Updated : 16 Aug, 2022

Two players are playing a game where string str is given. The first player can take the characters at even indices and the second player can take the characters at odd indices. The player which can build the lexicographically smaller string than the other player wins the game. Print the winner of the game, either player A, B or print Tie if it’s a tie.

Examples:

Input: str = “geeksforgeeks”
Output:
Explanation: “eeggoss” is the lexicographically smallest
string that player A can get.
“eefkkr” is the lexicographically smallest
string that player B can get.
And B’s string is lexicographically smaller.

Input: str = “abcdbh”
Output: A

Approach: Create two empty strings str1 and str2 for players A and B respectively. Traverse the original string character by character and for every character whose index is even, append this character in str1 else append this character in str2. Finally, sort the generated string in order to get the lexicographically smallest possible string and compare them to find the winner of the game.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to find the winner of the game` `void` `find_winner(string str, ``int` `n)` `{`   `    ``// To store the strings for both the players` `    ``string str1 = ``""``, str2 = ``""``;` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// If the index is even` `        ``if` `(i % 2 == 0) {`   `            ``// Append the current character` `            ``// to player A's string` `            ``str1 += str[i];` `        ``}`   `        ``// If the index is odd` `        ``else` `{`   `            ``// Append the current character` `            ``// to player B's string` `            ``str2 += str[i];` `        ``}` `    ``}`   `    ``// Sort both the strings to get` `    ``// the lexicographically smallest` `    ``// string possible` `    ``sort(str1.begin(), str1.end());` `    ``sort(str2.begin(), str2.end());`   `    ``// Compare both the strings to` `    ``// find the winner of the game` `    ``if` `(str1 < str2)` `        ``cout << ``"A"``;` `    ``else` `if` `(str2 < str1)` `        ``cout << ``"B"``;` `    ``else` `        ``cout << ``"Tie"``;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"geeksforgeeks"``;` `    ``int` `n = str.length();`   `    ``find_winner(str, n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.Arrays;`   `class` `GFG {`   `    ``// Function to find the winner of the game` `    ``static` `void` `find_winner(String str, ``int` `n)` `    ``{`   `        ``// To store the strings for both the players` `        ``String str1 = ``""``, str2 = ``""``;` `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// If the index is even` `            ``if` `(i % ``2` `== ``0``) {`   `                ``// Append the current character` `                ``// to player A's string` `                ``str1 += str.charAt(i);` `            ``}`   `            ``// If the index is odd` `            ``else` `{`   `                ``// Append the current character` `                ``// to player B's string` `                ``str2 += str.charAt(i);` `            ``}` `        ``}`   `        ``// Sort both the strings to get` `        ``// the lexicographically smallest` `        ``// string possible` `        ``char` `a[] = str1.toCharArray();` `        ``Arrays.sort(a);` `        ``char` `b[] = str2.toCharArray();` `        ``Arrays.sort(b);` `        ``str1 = ``new` `String(a);` `        ``str2 = ``new` `String(b);`   `        ``// Compare both the strings to` `        ``// find the winner of the game` `        ``if` `(str1.compareTo(str2) < ``0``)` `            ``System.out.print(``"A"``);` `        ``else` `if` `(str1.compareTo(str2) > ``0``)` `            ``System.out.print(``"B"``);` `        ``else` `            ``System.out.print(``"Tie"``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String str = ``"geeksforgeeks"``;` `        ``int` `n = str.length();`   `        ``find_winner(str, n);` `    ``}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach`   `# Function to find the winner of the game`     `def` `find_winner(string, n):`   `    ``# To store the strings for both the players` `    ``string1 ``=` `""` `    ``string2 ``=` `""` `    ``for` `i ``in` `range``(n):`   `        ``# If the index is even` `        ``if` `(i ``%` `2` `=``=` `0``):`   `            ``# Append the current character` `            ``# to player A's string` `            ``string1 ``+``=` `string[i]`   `        ``# If the index is odd` `        ``else``:`   `            ``# Append the current character` `            ``# to player B's string` `            ``string2 ``+``=` `string[i]`   `    ``# Sort both the strings to get` `    ``# the lexicographically smallest` `    ``# string possible` `    ``string1 ``=` `"".join(``sorted``(string1))` `    ``string2 ``=` `"".join(``sorted``(string2))`   `    ``# Compare both the strings to` `    ``# find the winner of the game` `    ``if` `(string1 < string2):` `        ``print``(``"A"``, end``=``"")`   `    ``elif` `(string2 < string1):` `        ``print``(``"B"``, end``=``"")`   `    ``else``:` `        ``print``(``"Tie"``, end``=``"")`     `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``string ``=` `"geeksforgeeks"` `    ``n ``=` `len``(string)`   `    ``find_winner(string, n)`   `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG {`   `    ``// Function to find the winner of the game` `    ``static` `void` `find_winner(String str, ``int` `n)` `    ``{`   `        ``// To store the strings for both the players` `        ``String str1 = ``""``, str2 = ``""``;` `        ``for` `(``int` `i = 0; i < n; i++) {`   `            ``// If the index is even` `            ``if` `(i % 2 == 0) {`   `                ``// Append the current character` `                ``// to player A's string` `                ``str1 += str[i];` `            ``}`   `            ``// If the index is odd` `            ``else` `{`   `                ``// Append the current character` `                ``// to player B's string` `                ``str2 += str[i];` `            ``}` `        ``}`   `        ``// Sort both the strings to get` `        ``// the lexicographically smallest` `        ``// string possible` `        ``char``[] a = str1.ToCharArray();` `        ``Array.Sort(a);` `        ``char``[] b = str2.ToCharArray();` `        ``Array.Sort(b);` `        ``str1 = ``new` `String(a);` `        ``str2 = ``new` `String(b);`   `        ``// Compare both the strings to` `        ``// find the winner of the game` `        ``if` `(str1.CompareTo(str2) < 0)` `            ``Console.Write(``"A"``);` `        ``else` `if` `(str1.CompareTo(str2) > 0)` `            ``Console.Write(``"B"``);` `        ``else` `            ``Console.Write(``"Tie"``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``String str = ``"geeksforgeeks"``;` `        ``int` `n = str.Length;`   `        ``find_winner(str, n);` `    ``}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`B`

Time Complexity: O(n log(n)), Where n is the length of the given string.
Auxiliary Space: O(n), for storing the strings for both the players.

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