Optimal Strategy for a Game | DP-31

Difficulty Level : Hard
Last Updated : 15 Apr, 2021

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:

5, 3, 7, 10 : The user collects maximum value as 15(10 + 5)
8, 15, 3, 7 : The user collects maximum value as 22(7 + 15)

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

…….User chooses 8.
…….Opponent chooses 15.
…….User chooses 7.
…….Opponent chooses 3.
Total value collected by user is 15(8 + 7)
…….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, the maximum value can be collected although the first move is not the best.

Approach: As both the players are equally strong, both will try to reduce the possibility of winning of each other. Now let’s see how the opponent can achieve this.

There are two choices:

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 + min(F(i+2, j), F(i+1, j-1) ).

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 + min(F(i+1, j-1), F(i, j-2) ).

Following is the recursive solution that is based on the above two choices. We take a maximum of two choices.

F(i, j) represents the maximum value the user
can collect from i'th coin to j'th coin.

F(i, j) = Max(Vi + min(F(i+2, j), F(i+1, j-1) ), 
              Vj + min(F(i+1, j-1), F(i, j-2) ))
As user wants to maximise the number of coins. 

Base Cases
    F(i, j) = Vi           If j == i
    F(i, j) = max(Vi, Vj)  If j == i + 1

C++

// C++ program to find out
// maximum value from a given
// sequence of coins
#include <bits/stdc++.h>
using namespace std;
 
// 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)
{
    // Create a table to store
    // solutions of subproblems
    int table[n][n];
 
    // Fill table using above
    // recursive formula. Note
    // that the table is filled
    // in diagonal fashion (similar
    // to http:// goo.gl/PQqoS),
    // from diagonal elements to
    // table[0][n-1] which is the result.
    for (int gap = 0; gap < n; ++gap) {
        for (int i = 0, j = gap; j < n; ++i, ++j) {
            // Here x is value of F(i+2, j),
            // y is F(i+1, j-1) and
            // z is F(i, j-2) in above recursive
            // formula
            int x = ((i + 2) <= j)
                        ? table[i + 2][j]
                        : 0;
            int y = ((i + 1) <= (j - 1))
                        ? table[i + 1][j - 1]
                        : 0;
            int z = (i <= (j - 2))
                        ? table[i][j - 2]
                        : 0;
 
            table[i][j] = max(
                arr[i] + min(x, y),
                arr[j] + min(y, z));
        }
    }
 
    return table[0][n - 1];
}
 
// Driver program to test above function
int 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

// Java program to find out maximum
// value from a given sequence of coins
import java.io.*;
 
class GFG {
 
    // 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)
    {
        // Create a table to store
        // solutions of subproblems
        int table[][] = new int[n][n];
        int gap, i, j, x, y, z;
 
        // Fill table using above recursive formula.
        // Note that the tableis filled in diagonal
        // fashion (similar to http:// goo.gl/PQqoS),
        // from diagonal elements to table[0][n-1]
        // which is the result.
        for (gap = 0; gap < n; ++gap) {
            for (i = 0, j = gap; j < n; ++i, ++j) {
 
                // Here x is value of F(i+2, j),
                // y is F(i+1, j-1) and z is
                // F(i, j-2) in above recursive formula
                x = ((i + 2) <= j)
                        ? table[i + 2][j]
                        : 0;
                y = ((i + 1) <= (j - 1))
                        ? table[i + 1][j - 1]
                        : 0;
                z = (i <= (j - 2))
                        ? table[i][j - 2]
                        : 0;
 
                table[i][j] = Math.max(
                    arr[i] + Math.min(x, y),
                    arr[j] + Math.min(y, z));
            }
        }
 
        return table[0][n - 1];
    }
 
    // Driver program
    public static 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 vt_m

Python3

# Python3 program to find out maximum
# value from a given sequence of coins
 
# 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):
     
    # Create a table to store
    # solutions of subproblems
    table = [[0 for i in range(n)]
                for i in range(n)]
 
    # Fill table using above recursive
    # formula. Note that the table is
    # filled in diagonal fashion
    # (similar to http://goo.gl / PQqoS),
    # from diagonal elements to
    # table[0][n-1] which is the result.
    for gap in range(n):
        for j in range(gap, n):
            i = j - gap
             
            # Here x is value of F(i + 2, j),
            # y is F(i + 1, j-1) and z is
            # F(i, j-2) in above recursive
            # formula
            x = 0
            if((i + 2) <= j):
                x = table[i + 2][j]
            y = 0
            if((i + 1) <= (j - 1)):
                y = table[i + 1][j - 1]
            z = 0
            if(i <= (j - 2)):
                z = table[i][j - 2]
            table[i][j] = max(arr[i] + min(x, y),
                              arr[j] + min(y, z))
    return table[0][n - 1]
 
# 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 contibuted
# by sahilshelangia

C#

// C# program to find out maximum
// value from a given sequence of coins
using System;
 
public class GFG {
 
    // 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)
    {
        // Create a table to store solutions of subproblems
        int[, ] table = new int[n, n];
        int gap, i, j, x, y, z;
 
        // Fill table using above recursive formula.
        // Note that the tableis filled in diagonal
        // fashion (similar to http:// goo.gl/PQqoS),
        // from diagonal elements to table[0][n-1]
        // which is the result.
        for (gap = 0; gap < n; ++gap) {
            for (i = 0, j = gap; j < n; ++i, ++j) {
 
                // Here x is value of F(i+2, j),
                // y is F(i+1, j-1) and z is
                // F(i, j-2) in above recursive formula
                x = ((i + 2) <= j) ? table[i + 2, j] : 0;
                y = ((i + 1) <= (j - 1)) ? table[i + 1, j - 1] : 0;
                z = (i <= (j - 2)) ? table[i, j - 2] : 0;
 
                table[i, j] = Math.Max(arr[i] + Math.Min(x, y),
                                       arr[j] + Math.Min(y, z));
            }
        }
 
        return table[0, n - 1];
    }
 
    // Driver program
 
    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 ajit

PHP

<?php
// PHP program to find out maximum value
// from a given sequence of coins
 
// 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)
{
    // Create a table to store solutions
    // of subproblems
    $table = array_fill(0, $n,
             array_fill(0, $n, 0));
 
    // Fill table using above recursive formula.
    // Note that the table is filled in diagonal
    // fashion (similar to http://goo.gl/PQqoS),
    // from diagonal elements to table[0][n-1]
    // which is the result.
    for ($gap = 0; $gap < $n; ++$gap)
    {
        for ($i = 0, $j = $gap; $j < $n; ++$i, ++$j)
        {
 
            // Here x is value of F(i+2, j),
            // y is F(i+1, j-1) and z is F(i, j-2)
            // in above recursive formula
            $x = (($i + 2) <= $j) ?
               $table[$i + 2][$j] : 0;
            $y = (($i + 1) <= ($j - 1)) ?
                 $table[$i + 1][$j - 1] : 0;
            $z = ($i <= ($j - 2)) ?
               $table[$i][$j - 2] : 0;
 
            $table[$i][$j] = max($arr[$i] + min($x, $y),
                                 $arr[$j] + min($y, $z));
        }
    }
 
    return $table[0][$n - 1];
}
 
// Driver Code
$arr1 = array( 8, 15, 3, 7 );
$n = count($arr1);
print(optimalStrategyOfGame($arr1, $n) . "\n");
 
$arr2 = array( 2, 2, 2, 2 );
$n = count($arr2);
print(optimalStrategyOfGame($arr2, $n) . "\n");
 
$arr3 = array(20, 30, 2, 2, 2, 10);
$n = count($arr3);
print(optimalStrategyOfGame($arr3, $n) . "\n");
 
// This code is contributed by chandan_jnu
?>

Javascript

<script>
 
// Javascript program to find out maximum
// value from a given sequence of coins
 
// 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)
{
     
    // Create a table to store
    // solutions of subproblems
    let table = new Array(n);
    let gap, i, j, x, y, z;
     
    for(let d = 0; d < n; d++)
    {
        table[d] = new Array(n);
    }
 
    // Fill table using above recursive formula.
    // Note that the tableis filled in diagonal
    // fashion (similar to http:// goo.gl/PQqoS),
    // from diagonal elements to table[0][n-1]
    // which is the result.
    for(gap = 0; gap < n; ++gap)
    {
        for(i = 0, j = gap; j < n; ++i, ++j)
        {
             
            // Here x is value of F(i+2, j),
            // y is F(i+1, j-1) and z is
            // F(i, j-2) in above recursive formula
            x = ((i + 2) <= j) ? table[i + 2][j] : 0;
            y = ((i + 1) <= (j - 1)) ?
            table[i + 1][j - 1] : 0;
            z = (i <= (j - 2)) ? table[i][j - 2] : 0;
 
            table[i][j] = Math.max(
                arr[i] + Math.min(x, y),
                arr[j] + Math.min(y, z));
        }
    }
    return table[0][n - 1];
}
 
// Driver code
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));
   
// This code is contributed by divyesh072019
 
</script>

Output:

22
4
42

Complexity Analysis:

Time Complexity: O(n²).
Use of a nested for loop brings the time complexity to n².
Auxiliary Space: O(n²).
As a 2-D table is used for storing states.

Note: The above solution can be optimized by using less number of comparisons for every choice. Please refer below.
Optimal Strategy for a Game | Set 2
Exercise:
Your thoughts on the strategy when the user wishes to only win instead of winning with the maximum value. Like the above problem, the number of coins is even.
Can the Greedy approach work quite well and give an optimal solution? Will your answer change if the number of coins is odd? Please see Coin game of two corners
This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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