Find the winner of the Game (original) (raw)

Last Updated : 12 Jul, 2025

Given an array arr[] of N integers and two players A and B are playing a game where the players pick the element with the maximum digit sum in their turns. In the end, the player with the maximum sum of the picked elements wins the game. Assuming that player A always starts the game first and both the players play optimally, the task is to find the winner of the game.
Examples:

Input: arr[] = {12, 43, 25, 23, 30}
Output: B
A choses 43
B chooses 25
A chooses 23
B chooses 30
A chooses 12
A's score = 43 + 23 + 12 = 78
B's score = 25 + 30 = 55
Input: arr[] = {2, 1, 1, 2}
Output: Draw

Approach: Sort the array based on the digit sum values of the integers, if the digit sum of two integers is same then they will be compared based on their values, this is because the value will maximize the sum in the end. After the array has been sorted based on the custom comparator, player A will try to pick the elements starting from the greatest (greedily).
Below is the implementation of the above approach:

C++ `

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;

// Function that returns the // sum of the digits of n int digit_sum(int n) { int s = 0; while (n > 0) { s += n % 10; n /= 10; } return s; }

// Compares two integers according // to their digit sum bool comparator(int a, int b) {

// Sum of digits of a
int s1 = digit_sum(a);

// Sum of digits of b
int s2 = digit_sum(b);

// If the digit sum of a is equal
// to the digit sum of b
if (s1 == s2)
    return (a < b);
return (s1 < s2);

}

// Function to return the winner of the game string findTheWinner(int arr[], int n) {

// Sort the elements based on
// the digit sum values
sort(arr, arr + n, comparator);

// Find player A's score
int scoreA = 0;
for (int i = n - 1; i >= 0; i -= 2)
    scoreA += arr[i];

// Find player A's score
int scoreB = 0;
for (int i = n - 2; i >= 0; i -= 2)
    scoreB += arr[i];

// Find the winner
if (scoreA == scoreB)
    return "Draw";
else if (scoreA > scoreB)
    return "A";
return "B";

}

// Driver code int main() { int arr[] = { 12, 43, 25, 23, 30 }; int n = sizeof(arr) / sizeof(int);

cout << findTheWinner(arr, n);

return 0;

}

Java

/*package whatever //do not write package name here / import java.util.; public class GFG {

// Function that returns the // sum of the digits of n static int digit_sum(int n) { int s = 0; while (n > 0) { s += n % 10; n /= 10; } return s; }

// Function to return the winner of the game static String findTheWinner(int arr[], int n) {

int arr1[][] = new int[arr.length][2];

for(int i=0;i<arr.length;i++){
  arr1[i][0] = arr[i];
  arr1[i][1] = digit_sum(arr[i]);
}

// Sort the elements based on
// the digit sum values
Arrays.sort(arr1,(a,b)->a[0]-b[0]);

// Find player A's score
int scoreA = 0;
for (int i = n - 1; i >= 0; i -= 2)
  scoreA += arr1[i][0];

// Find player A's score
int scoreB = 0;
for (int i = n - 2; i >= 0; i -= 2)
  scoreB += arr1[i][0];

// Find the winner
if (scoreA == scoreB)
  return "Draw";
else if (scoreA > scoreB)
  return "A";
return "B";

}

public static void main (String[] args) { int arr[] = { 12, 43, 25, 23, 30 }; int n = arr.length;

System.out.println(findTheWinner(arr, n));

} }

// This code is contributed by aadityaburujwale.

Python3

Python3 implementation of the above approach

Function that returns the

sum of the digits of n

def digit_sum(n):

s = 0; 
while n > 0: 
    s += n % 10
    n /= 10
    
return s

Function to return the winner

of the game

def findTheWinner(arr, n):

# Sort the elements based on 
# the digit sum values 
arr.sort(key = digit_sum)

# Find player A's score 
scoreA = 0
i = n - 1
while i >= 0:
    scoreA += arr[i]
    i -= 2

# Find player A's score 
scoreB = 0
i = n - 2
while i >= 0:
    scoreA += arr[i]
    i -= 2

# Find the winner 
if scoreA == scoreB: 
    return "Draw"
elif (scoreA > scoreB): 
    return "A"
    
return "B"

Driver code

if name=="main":

arr = [ 12, 43, 25, 23, 30 ]
n = len(arr); 

print(findTheWinner(arr, n)) 

This code is contributed by Yash_R

C#

// C# implementation of the approach using System; class GFG {

// Function that returns the // sum of the digits of n static int digit_sum(int n) { int s = 0; while (n > 0) { s += n % 10; n /= 10; } return s; }

// Compares two integers according // to their digit sum static bool comparator(int a, int b) {

// Sum of digits of a
int s1 = digit_sum(a);

// Sum of digits of b
int s2 = digit_sum(b);

// If the digit sum of a is equal
// to the digit sum of b
if (s1 == s2)
  return (a < b);
return (s1 < s2);

} // Function to return the winner of the game static string findTheWinner(int []arr, int n) {

// Sort the elements based on
// the digit sum values
// sort(arr, arr + n, comparator);
Array.Sort(arr,new Comparison<int>(
  (i1, i2) =>  digit_sum(i2).CompareTo( digit_sum(i1))));
// Find player A's score
int scoreA = 0;
for (int i = n - 1; i >= 0; i -= 2)
  scoreA += arr[i];

// Find player A's score
int scoreB = 0;
for (int i = n - 2; i >= 0; i -= 2)
  scoreB += arr[i];

// Find the winner
if (scoreA == scoreB)
  return "Draw";
else if (scoreA > scoreB)
  return "A";
return "B";

} static void Main() { int []arr = { 12, 43, 25, 23, 30 }; int n = arr.Length;

Console.Write(findTheWinner(arr, n));

} }

// This code is contributed by garg28harsh.

JavaScript

`

Time Complexity: O(n*log(n)*d) where n is the size of the array and d is the maximum number of digits in any number present in the array.

Auxiliary Space: O(1)