Optimal Strategy for a Game | Set 2

• Difficulty Level : Hard
• Last Updated : 06 Jun, 2021

Problem statement: Consider a row of n coins of values v1 . . . vn, where n is even. We play a game against an opponent by alternating turns. In each turn, a player selects either the first or last coin from the row, removes it from the row permanently, and receives the value of the coin. Determine the maximum possible amount of money we can definitely win if we move first.
Note: The opponent is as clever as the user.

Let us understand the problem with few examples:
1. 5, 3, 7, 10 : The user collects maximum value as 15(10 + 5)
2. 8, 15, 3, 7 : The user collects maximum value as 22(7 + 15)

Does choosing the best at each move give an optimal solution?
No. In the second example, this is how the game can finish:

1.
…….User chooses 8.
…….Opponent chooses 15.
…….User chooses 7.
…….Opponent chooses 3.
Total value collected by user is 15(8 + 7)
2.
…….User chooses 7.
…….Opponent chooses 8.
…….User chooses 15.
…….Opponent chooses 3.
Total value collected by user is 22(7 + 15)
So if the user follows the second game state, maximum value can be collected although the first move is not the best.

We have discussed an approach that makes 4 recursive calls. In this post, a new approach is discussed that makes two recursive calls.
There are two choices:
1. The user chooses the ith coin with value Vi: The opponent either chooses (i+1)th coin or jth coin. The opponent intends to choose the coin which leaves the user with minimum value.
i.e. The user can collect the value Vi + (Sum – Vi) – F(i+1, j, Sum – Vi) where Sum is sum of coins from index i to j. The expression can be simplified to Sum – F(i+1, j, Sum – Vi) 2. The user chooses the jth coin with value Vj: The opponent either chooses ith coin or (j-1)th coin. The opponent intends to choose the coin which leaves the user with minimum value.
i.e. The user can collect the value Vj + (Sum – Vj) – F(i, j-1, Sum – Vj) where Sum is sum of coins from index i to j. The expression can be simplified to Sum – F(i, j-1, Sum – Vj) Following is recursive solution that is based on above two choices. We take the maximum of two choices.

F(i, j)  represents the maximum value the user can collect from
i'th coin to j'th coin.
arr[]   represents the list of coins

F(i, j)  = Max(Sum - F(i+1, j, Sum-arr[i]),
Sum - F(i, j-1, Sum-arr[j]))
Base Case
F(i, j)  = max(arr[i], arr[j])  If j == i+1

Simple Recursive Solution

C++

 // C++ program to find out maximum value from a // given sequence of coins #include using namespace std;   int oSRec(int arr[], int i, int j, int sum) {     if (j == i + 1)         return max(arr[i], arr[j]);       // For both of your choices, the opponent     // gives you total sum minus maximum of     // his value     return max((sum - oSRec(arr, i + 1, j, sum - arr[i])),                (sum - oSRec(arr, i, j - 1, sum - arr[j]))); }   // Returns optimal value possible that a player can // collect from an array of coins of size n. Note // than n must be even int optimalStrategyOfGame(int* arr, int n) {     int sum = 0;     sum = accumulate(arr, arr + n, sum);     return oSRec(arr, 0, n - 1, sum); }   // Driver program to test above function int main() {     int arr1[] = { 8, 15, 3, 7 };     int n = sizeof(arr1) / sizeof(arr1);     printf("%d\n", optimalStrategyOfGame(arr1, n));       int arr2[] = { 2, 2, 2, 2 };     n = sizeof(arr2) / sizeof(arr2);     printf("%d\n", optimalStrategyOfGame(arr2, n));       int arr3[] = { 20, 30, 2, 2, 2, 10 };     n = sizeof(arr3) / sizeof(arr3);     printf("%d\n", optimalStrategyOfGame(arr3, n));       return 0; }

Java

 // Java program to find out maximum value from a // given sequence of coins import java .io.*;   class GFG {       static int oSRec(int []arr, int i, int j, int sum)     {         if (j == i + 1)             return Math.max(arr[i], arr[j]);               // For both of your choices, the opponent         // gives you total sum minus maximum of         // his value         return Math.max((sum - oSRec(arr, i + 1, j, sum - arr[i])),                 (sum - oSRec(arr, i, j - 1, sum - arr[j])));     }           // Returns optimal value possible that a player can     // collect from an array of coins of size n. Note     // than n must be even     static int optimalStrategyOfGame(int[] arr, int n)     {         int sum = 0;         for(int i = 0; i < n; i++)         {             sum += arr[i];         }           return oSRec(arr, 0, n - 1, sum);     }           // Driver code     static public void main (String[] args)     {         int []arr1 = { 8, 15, 3, 7 };         int n = arr1.length;         System.out.println(optimalStrategyOfGame(arr1, n));               int []arr2 = { 2, 2, 2, 2 };         n = arr2.length;         System.out.println(optimalStrategyOfGame(arr2, n));               int []arr3 = { 20, 30, 2, 2, 2, 10 };         n = arr3.length ;         System.out.println(optimalStrategyOfGame(arr3, n));     } }   // This code is contributed by anuj_67..

Python3

 # python3 program to find out maximum value from a # given sequence of coins   def oSRec (arr, i, j, Sum):       if (j == i + 1):         return max(arr[i], arr[j])       # For both of your choices, the opponent     # gives you total Sum minus maximum of     # his value     return max((Sum - oSRec(arr, i + 1, j, Sum - arr[i])),                 (Sum - oSRec(arr, i, j - 1, Sum - arr[j])))   # Returns optimal value possible that a player can # collect from an array of coins of size n. Note # than n must be even def optimalStrategyOfGame(arr, n):       Sum = 0     Sum = sum(arr)     return oSRec(arr, 0, n - 1, Sum)   # Driver code   arr1= [ 8, 15, 3, 7] n = len(arr1) print(optimalStrategyOfGame(arr1, n))   arr2= [ 2, 2, 2, 2 ] n = len(arr2) print(optimalStrategyOfGame(arr2, n))   arr3= [ 20, 30, 2, 2, 2, 10 ] n = len(arr3) print(optimalStrategyOfGame(arr3, n))   # This code is contributed by Mohit kumar 29

C#

 // C# program to find out maximum value from a // given sequence of coins using System; class GFG {     static int oSRec(int []arr, int i,                      int j, int sum)     {         if (j == i + 1)             return Math.Max(arr[i], arr[j]);               // For both of your choices, the opponent         // gives you total sum minus maximum of         // his value         return Math.Max((sum - oSRec(arr, i + 1, j,                                      sum - arr[i])),                         (sum - oSRec(arr, i, j - 1,                                      sum - arr[j])));     }           // Returns optimal value possible that a player can     // collect from an array of coins of size n. Note     // than n must be even     static int optimalStrategyOfGame(int[] arr, int n)     {         int sum = 0;         for(int i = 0; i < n; i++)         {             sum += arr[i];         }           return oSRec(arr, 0, n - 1, sum);     }           // Driver code     static public void Main ()     {         int []arr1 = { 8, 15, 3, 7 };         int n = arr1.Length;         Console.WriteLine(optimalStrategyOfGame(arr1, n));               int []arr2 = { 2, 2, 2, 2 };         n = arr2.Length;         Console.WriteLine(optimalStrategyOfGame(arr2, n));               int []arr3 = { 20, 30, 2, 2, 2, 10 };         n = arr3.Length;         Console.WriteLine(optimalStrategyOfGame(arr3, n));     } }   // This code is contributed by AnkitRai01

Javascript



Output:

22
4
42

Memoization Based Solution

C++

 // C++ program to find out maximum value from a // given sequence of coins #include using namespace std;   const int MAX = 100;   int memo[MAX][MAX];   int oSRec(int arr[], int i, int j, int sum) {     if (j == i + 1)         return max(arr[i], arr[j]);       if (memo[i][j] != -1)         return memo[i][j];       // For both of your choices, the opponent     // gives you total sum minus maximum of     // his value     memo[i][j] = max((sum - oSRec(arr, i + 1, j, sum - arr[i])),                      (sum - oSRec(arr, i, j - 1, sum - arr[j])));       return memo[i][j]; }   // Returns optimal value possible that a player can // collect from an array of coins of size n. Note // than n must be even int optimalStrategyOfGame(int* arr, int n) {     // Compute sum of elements     int sum = 0;     sum = accumulate(arr, arr + n, sum);       // Initialize memoization table     memset(memo, -1, sizeof(memo));       return oSRec(arr, 0, n - 1, sum); }   // Driver program to test above function int main() {     int arr1[] = { 8, 15, 3, 7 };     int n = sizeof(arr1) / sizeof(arr1);     printf("%d\n", optimalStrategyOfGame(arr1, n));       int arr2[] = { 2, 2, 2, 2 };     n = sizeof(arr2) / sizeof(arr2);     printf("%d\n", optimalStrategyOfGame(arr2, n));       int arr3[] = { 20, 30, 2, 2, 2, 10 };     n = sizeof(arr3) / sizeof(arr3);     printf("%d\n", optimalStrategyOfGame(arr3, n));       return 0; }

Java

 // Java program to find out maximum value from a // given sequence of coins import java.util.*; class GFG{   static int MAX = 100;   static int [][]memo = new int[MAX][MAX];   static int oSRec(int arr[], int i,                  int j, int sum) {     if (j == i + 1)         return Math.max(arr[i], arr[j]);       if (memo[i][j] != -1)         return memo[i][j];       // For both of your choices, the opponent     // gives you total sum minus maximum of     // his value     memo[i][j] = Math.max((sum - oSRec(arr, i + 1, j,                                        sum - arr[i])),                            (sum - oSRec(arr, i, j - 1,                                        sum - arr[j])));       return memo[i][j]; }   static int accumulate(int[] arr,                       int start, int end) {     int sum=0;     for(int i= 0; i < arr.length; i++)         sum += arr[i];     return sum; }     // Returns optimal value possible that a player can // collect from an array of coins of size n. Note // than n must be even static int optimalStrategyOfGame(int []arr, int n) {     // Compute sum of elements     int sum = 0;     sum = accumulate(arr, 0, n);       // Initialize memoization table     for (int j = 0; j < MAX; j++)     {         for (int k = 0; k < MAX; k++)             memo[j][k] = -1;     }       return oSRec(arr, 0, n - 1, sum); }   // Driver Code public static void main(String[] args) {     int arr1[] = { 8, 15, 3, 7 };     int n = arr1.length;     System.out.printf("%d\n",                optimalStrategyOfGame(arr1, n));       int arr2[] = { 2, 2, 2, 2 };     n = arr2.length;     System.out.printf("%d\n",                optimalStrategyOfGame(arr2, n));       int arr3[] = { 20, 30, 2, 2, 2, 10 };     n = arr3.length;     System.out.printf("%d\n",                optimalStrategyOfGame(arr3, n)); } }   // This code is contributed by gauravrajput1

Python3

 # Python3 program to find out maximum value # from a given sequence of coins MAX = 100    memo = [[0 for i in range(MAX)]            for j in range(MAX)]    def oSRec(arr, i, j, Sum):       if (j == i + 1):         return max(arr[i], arr[j])        if (memo[i][j] != -1):         return memo[i][j]        # For both of your choices, the opponent     # gives you total sum minus maximum of     # his value     memo[i][j] = max((Sum - oSRec(arr, i + 1, j,                                      Sum - arr[i])),                      (Sum - oSRec(arr, i, j - 1,                                         Sum - arr[j])))        return memo[i][j]      # Returns optimal value possible that a # player can collect from an array of # coins of size n. Note than n must # be even def optimalStrategyOfGame(arr, n):       # Compute sum of elements     Sum = 0     Sum = sum(arr)        # Initialize memoization table     for j in range(MAX):         for k in range(MAX):             memo[j][k] = -1        return oSRec(arr, 0, n - 1, Sum)    # Driver Code arr1 = [ 8, 15, 3, 7 ] n = len(arr1) print(optimalStrategyOfGame(arr1, n))   arr2 = [ 2, 2, 2, 2 ] n = len(arr2) print(optimalStrategyOfGame(arr2, n))   arr3 = [ 20, 30, 2, 2, 2, 10 ] n = len(arr3) print(optimalStrategyOfGame(arr3, n))   # This code is contributed by divyesh072019

C#

 // C# program to find out maximum value from a // given sequence of coins using System; class GFG{     static int MAX = 100;     static int[,] memo = new int[MAX, MAX];     static int oSRec(int []arr, int i,                    int j, int sum)   {     if (j == i + 1)       return Math.Max(arr[i], arr[j]);       if (memo[i, j] != -1)       return memo[i, j];       // For both of your choices, the opponent     // gives you total sum minus maximum of     // his value     memo[i, j] = Math.Max((sum - oSRec(arr, i + 1, j,                                        sum - arr[i])),                           (sum - oSRec(arr, i, j - 1,                                        sum - arr[j])));       return memo[i,j];   }     static int accumulate(int[] arr, int start,                         int end)   {     int sum = 0;     for (int i = 0; i < arr.Length; i++)       sum += arr[i];     return sum;   }     // Returns optimal value possible that a player can   // collect from an array of coins of size n. Note   // than n must be even   static int optimalStrategyOfGame(int[] arr, int n)   {     // Compute sum of elements     int sum = 0;     sum = accumulate(arr, 0, n);       // Initialize memoization table     for (int j = 0; j < MAX; j++)     {       for (int k = 0; k < MAX; k++)         memo[j, k] = -1;     }       return oSRec(arr, 0, n - 1, sum);   }     // Driver Code   public static void Main(String[] args)   {     int []arr1 = { 8, 15, 3, 7 };     int n = arr1.Length;     Console.Write("{0}\n", optimalStrategyOfGame(arr1, n));       int []arr2 = { 2, 2, 2, 2 };     n = arr2.Length;     Console.Write("{0}\n", optimalStrategyOfGame(arr2, n));       int []arr3 = { 20, 30, 2, 2, 2, 10 };     n = arr3.Length;     Console.Write("{0}\n", optimalStrategyOfGame(arr3, n));   } }   // This code is contributed by Rohit_ranjan

Javascript



Output:

22
4
42

This approach is suggested by Alind.

My Personal Notes arrow_drop_up
Recommended Articles
Page :