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

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  • Last Updated : 16 Aug, 2022
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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 <bits/stdc++.h>
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




<script>
 
// Javascript implementation of the approach
 
// Function to find the winner of the game
function find_winner(str, n)
{
 
    // To store the strings for both the players
    var str1 = "", str2 = "";
    for (var 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
    str1 = str1.split('').sort();
    str2 = str2.split('').sort();
 
    // Compare both the strings to
    // find the winner of the game
    if (str1 < str2)
        document.write( "A");
    else if (str2 < str1)
        document.write( "B");
    else
        document.write( "Tie");
}
 
// Driver code
var str = "geeksforgeeks";
var n = str.length;
find_winner(str, n);
 
</script>


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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