Optimal Strategy for a Game | Set 2

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++
Java
Python3
C#
Javascript
// C++ program to find out maximum value from a// given sequence of coins#include <bits/stdc++.h>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 evenint 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 functionint main(){    int arr1[] = { 8, 15, 3, 7 };    int n = sizeof(arr1) / sizeof(arr1[0]);    printf("%d\n", optimalStrategyOfGame(arr1, n));     int arr2[] = { 2, 2, 2, 2 };    n = sizeof(arr2) / sizeof(arr2[0]);    printf("%d\n", optimalStrategyOfGame(arr2, n));     int arr3[] = { 20, 30, 2, 2, 2, 10 };    n = sizeof(arr3) / sizeof(arr3[0]);    printf("%d\n", optimalStrategyOfGame(arr3, n));     return 0;}
// Java program to find out maximum value from a// given sequence of coinsimport 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 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 evendef 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# program to find out maximum value from a// given sequence of coinsusing 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
<script>    // Javascript program to find out maximum value from a    // given sequence of coins         function oSRec(arr, i, j, 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    function optimalStrategyOfGame(arr, n)    {        let sum = 0;        for(let i = 0; i < n; i++)        {            sum += arr[i];        }          return oSRec(arr, 0, n - 1, sum);    }         let arr1 = [ 8, 15, 3, 7 ];    let n = arr1.length;    document.write(optimalStrategyOfGame(arr1, n) + "</br>");     let arr2 = [ 2, 2, 2, 2 ];    n = arr2.length;    document.write(optimalStrategyOfGame(arr2, n) + "</br>");     let arr3 = [ 20, 30, 2, 2, 2, 10 ];    n = arr3.length;    document.write(optimalStrategyOfGame(arr3, n) + "</br>"); </script>
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Memoization Based Solution 
C++
Java
Python3
C#
Javascript
// C++ program to find out maximum value from a// given sequence of coins#include <bits/stdc++.h>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 evenint 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 functionint main(){    int arr1[] = { 8, 15, 3, 7 };    int n = sizeof(arr1) / sizeof(arr1[0]);    printf("%d\n", optimalStrategyOfGame(arr1, n));     int arr2[] = { 2, 2, 2, 2 };    n = sizeof(arr2) / sizeof(arr2[0]);    printf("%d\n", optimalStrategyOfGame(arr2, n));     int arr3[] = { 20, 30, 2, 2, 2, 10 };    n = sizeof(arr3) / sizeof(arr3[0]);    printf("%d\n", optimalStrategyOfGame(arr3, n));     return 0;}
// Java program to find out maximum value from a// given sequence of coinsimport 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 evenstatic 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 Codepublic 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 program to find out maximum value# from a given sequence of coinsMAX = 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 evendef 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 Codearr1 = [ 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# program to find out maximum value from a// given sequence of coinsusing 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
<script> // Javascript program to find out maximum// value from a given sequence of coinslet MAX = 100;let memo = new Array(MAX);for(let i = 0; i < MAX; i++){    memo[i] = new Array(MAX);    for(let j = 0; j < MAX; j++)    {        memo[i][j] = 0;    }} function oSRec(arr, i, j, 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];} function accumulate(arr, start, end){     let sum = 0;    for(let 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 evenfunction optimalStrategyOfGame(arr, n){         // Compute sum of elements    let sum = 0;    sum = accumulate(arr, 0, n);      // Initialize memoization table    for(let j = 0; j < MAX; j++)    {        for(let k = 0; k < MAX; k++)            memo[j][k] = -1;    }    return oSRec(arr, 0, n - 1, sum);} // Driver Codelet arr1 = [ 8, 15, 3, 7 ];let n = arr1.length;document.write(    optimalStrategyOfGame(arr1, n) + "<br>");                let arr2 = [ 2, 2, 2, 2 ];n = arr2.length;document.write(    optimalStrategyOfGame(arr2, n) + "<br>");                let arr3 = [ 20, 30, 2, 2, 2, 10 ];n = arr3.length;document.write(    optimalStrategyOfGame(arr3, n) + "<br>"); // This code is contributed by patel2127 </script>
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This approach is suggested by Alind. 
mohit kumar 29
ankthon
vt_m
ShrabanaBiswas
zeeshan amber
GauravRajput1
Rohit_ranjan
divyesh072019
decode2207
patel2127
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