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++ 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;
} |
/*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 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# 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. |
<script> // Javascript implementation of the approach // Function that returns the // sum of the digits of n function digit_sum(n)
{ var s = 0;
while (n > 0) {
s += n % 10;
n = parseInt(n/10);
}
return s;
} // Compares two integers according // to their digit sum function comparator(a, b)
{ // Sum of digits of a
var s1 = digit_sum(a);
// Sum of digits of b
var 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 function findTheWinner(arr, n)
{ // Sort the elements based on
// the digit sum values
arr.sort(comparator);
// Find player A's score
var scoreA = 0;
for ( var i = n - 1; i >= 0; i -= 2)
scoreA += arr[i];
// Find player A's score
var scoreB = 0;
for ( var 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 var arr = [ 12, 43, 25, 23, 30 ];
var n = arr.length;
document.write( findTheWinner(arr, n)); </script> |
A
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)