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.
Examples:
Input: {5, 3, 7, 10}
Output: 15 -> (10 + 5)Input: {8, 15, 3, 7}
Output: 22 -> (7 + 15)
Why greedy algorithm fails here?
Does choosing the best at each move give an optimal solution? No.
In the second example, this is how the game can be finished in two ways:
- …….The user chooses 8.
…….The opponent chooses 15.
…….The user chooses 7.
…….The opponent chooses 3.
The total value collected by the user is 15(8 + 7) - …….The user chooses 7.
…….The opponent chooses 8.
…….The user chooses 15.
…….The opponent chooses 3.
The total value collected by the user is 22(7 + 15)
Note: If the user follows the second game state, the maximum value can be collected although the first move is not the best.
Optimal Strategy for a Game using memoization:
To solve the problem follow the below idea:
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) ) where [i+2,j] is the range of array indices available to the user if the opponent chooses Vi+1 and [i+1,j-1] is the range of array indexes available if opponent chooses the jth coin.
- 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 the minimum value, i.e. the user can collect the value Vj + min(F(i+1, j-1), F(i, j-2) ) where [i,j-2] is the range of array indices available for the user if the opponent picks jth coin and [i+1,j-1] is the range of indices available to the user if the opponent picks up the ith coin.
Below is the recursive approach 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
Below is the implementation of the above approach:
C++
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std; vector<int> arr; map<vector<int>, int> memo; int n = arr.size(); // recursive top down memoized solution int solve(int i, int j) { if ((i > j) || (i >= n) || (j < 0)) return 0; vector<int> k{ i, j }; if (memo[k] != 0) return memo[k]; // if the user chooses ith coin, the opponent can choose // from i+1th or jth coin. if he chooses i+1th coin, // user is left with [i+2,j] range. if opp chooses jth // coin, then user is left with [i+1,j-1] range to // choose from. Also opponent tries to choose in such a // way that the user has minimum value left. int option1 = arr[i] + min(solve(i + 2, j), solve(i + 1, j - 1)); // if user chooses jth coin, opponent can choose ith // coin or j-1th coin. if opp chooses ith coin,user can // choose in range [i+1,j-1]. if opp chooses j-1th coin, // user can choose in range [i,j-2]. int option2 = arr[j] + min(solve(i + 1, j - 1), solve(i, j - 2)); // since the user wants to get maximum money memo[k] = max(option1, option2); return memo[k]; } int optimalStrategyOfGame() { memo.clear(); return solve(0, n - 1); } // Driver code int main() { arr.push_back(8); arr.push_back(15); arr.push_back(3); arr.push_back(7); n = arr.size(); cout << optimalStrategyOfGame() << endl; arr.clear(); arr.push_back(2); arr.push_back(2); arr.push_back(2); arr.push_back(2); n = arr.size(); cout << optimalStrategyOfGame() << endl; arr.clear(); arr.push_back(20); arr.push_back(30); arr.push_back(2); arr.push_back(2); arr.push_back(2); arr.push_back(10); n = arr.size(); cout << optimalStrategyOfGame() << endl; } // This code is contributed by phasing17 |
Java
// Java code to implement the approach import java.util.*; class GFG { static ArrayList<Integer> arr = new ArrayList<>(); static HashMap<ArrayList<Integer>, Integer> memo = new HashMap<>(); static int n = 0; // recursive top down memoized solution static int solve(int i, int j) { if ((i > j) || (i >= n) || (j < 0)) return 0; ArrayList<Integer> k = new ArrayList<Integer>(); k.add(i); k.add(j); if (memo.containsKey(k)) return memo.get(k); // if the user chooses ith coin, the opponent can // choose from i+1th or jth coin. if he chooses // i+1th coin, user is left with [i+2,j] range. if // opp chooses jth coin, then user is left with // [i+1,j-1] range to choose from. Also opponent // tries to choose in such a way that the user has // minimum value left. int option1 = arr.get(i) + Math.min(solve(i + 2, j), solve(i + 1, j - 1)); // if user chooses jth coin, opponent can choose ith // coin or j-1th coin. if opp chooses ith coin,user // can choose in range [i+1,j-1]. if opp chooses // j-1th coin, user can choose in range [i,j-2]. int option2 = arr.get(j) + Math.min(solve(i + 1, j - 1), solve(i, j - 2)); // since the user wants to get maximum money memo.put(k, Math.max(option1, option2)); return memo.get(k); } static int optimalStrategyOfGame() { memo.clear(); return solve(0, n - 1); } // Driver code public static void main(String[] args) { arr.add(8); arr.add(15); arr.add(3); arr.add(7); n = arr.size(); System.out.println(optimalStrategyOfGame()); arr.clear(); arr.add(2); arr.add(2); arr.add(2); arr.add(2); n = arr.size(); System.out.println(optimalStrategyOfGame()); arr.clear(); arr.add(20); arr.add(30); arr.add(2); arr.add(2); arr.add(2); arr.add(10); n = arr.size(); System.out.println(optimalStrategyOfGame()); } } // This code is contributed by phasing17 |
Python3
# Python3 code to implement the approach def optimalStrategyOfGame(arr, n): memo = {} # recursive top down memoized solution def solve(i, j): if i > j or i >= n or j < 0: return 0 k = (i, j) if k in memo: return memo[k] # if the user chooses ith coin, the opponent can choose from i+1th or jth coin. # if he chooses i+1th coin, user is left with [i+2,j] range. # if opp chooses jth coin, then user is left with [i+1,j-1] range to choose from. # Also opponent tries to choose # in such a way that the user has minimum value left. option1 = arr[i] + min(solve(i+2, j), solve(i+1, j-1)) # if user chooses jth coin, opponent can choose ith coin or j-1th coin. # if opp chooses ith coin,user can choose in range [i+1,j-1]. # if opp chooses j-1th coin, user can choose in range [i,j-2]. option2 = arr[j] + min(solve(i+1, j-1), solve(i, j-2)) # since the user wants to get maximum money memo[k] = max(option1, option2) return memo[k] return solve(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 contributed # by sahilshelangia |
C#
// C# code to implement the approach using System; using System.Collections.Generic; class GFG { static List<int> arr = new List<int>(); static Dictionary<List<int>, int> memo = new Dictionary<List<int>, int>(); static int n = 0; // recursive top down memoized solution static int solve(int i, int j) { if ((i > j) || (i >= n) || (j < 0)) return 0; List<int> k = new List<int>{ i, j }; if (memo.ContainsKey(k)) return memo[k]; // if the user chooses ith coin, the opponent can // choose from i+1th or jth coin. if he chooses // i+1th coin, user is left with [i+2,j] range. if // opp chooses jth coin, then user is left with // [i+1,j-1] range to choose from. Also opponent // tries to choose in such a way that the user has // minimum value left. int option1 = arr[i] + Math.Min(solve(i + 2, j), solve(i + 1, j - 1)); // if user chooses jth coin, opponent can choose ith // coin or j-1th coin. if opp chooses ith coin,user // can choose in range [i+1,j-1]. if opp chooses // j-1th coin, user can choose in range [i,j-2]. int option2 = arr[j] + Math.Min(solve(i + 1, j - 1), solve(i, j - 2)); // since the user wants to get maximum money memo[k] = Math.Max(option1, option2); return memo[k]; } static int optimalStrategyOfGame() { memo.Clear(); return solve(0, n - 1); } // Driver code public static void Main(string[] args) { arr.Add(8); arr.Add(15); arr.Add(3); arr.Add(7); n = arr.Count; Console.WriteLine(optimalStrategyOfGame()); arr.Clear(); arr.Add(2); arr.Add(2); arr.Add(2); arr.Add(2); n = arr.Count; Console.WriteLine(optimalStrategyOfGame()); arr.Clear(); arr.Add(20); arr.Add(30); arr.Add(2); arr.Add(2); arr.Add(2); arr.Add(10); n = arr.Count; Console.WriteLine(optimalStrategyOfGame()); } } // This code is contributed by phasing17 |
Javascript
// JavaScript code to implement the approach <script> function optimalStrategyOfGame(arr, n) { let memo = {}; // recursive top down memoized solution function solve(i, j) { if ( (i > j) || (i >= n) || (j < 0)) return 0; let k = (i, j); if (memo.hasOwnProperty(k)) return memo[k]; // if the user chooses ith coin, the opponent can choose from i+1th or jth coin. // if he chooses i+1th coin, user is left with [i+2,j] range. // if opp chooses jth coin, then user is left with [i+1,j-1] range to choose from. // Also opponent tries to choose // in such a way that the user has minimum value left. let option1 = arr[i] + Math.min(solve(i+2, j), solve(i+1, j-1)); // if user chooses jth coin, opponent can choose ith coin or j-1th coin. // if opp chooses ith coin,user can choose in range [i+1,j-1]. // if opp chooses j-1th coin, user can choose in range [i,j-2]. let option2 = arr[j] + Math.min(solve(i+1, j-1), solve(i, j-2)); // since the user wants to get maximum money memo[k] = Math.max(option1, option2); return memo[k]; } return solve(0, n-1); } // Driver code let arr1 = [ 8, 15, 3, 7 ]; let n = arr1.length; console.log(optimalStrategyOfGame(arr1, n)); let arr2 = [ 2, 2, 2, 2 ]; n = arr2.length; console.log(optimalStrategyOfGame(arr2, n)); let arr3 = [ 20, 30, 2, 2, 2, 10 ]; n = arr3.length; console.log(optimalStrategyOfGame(arr3, n)); </script> // This code is contributed by phasing17 |
Output
22 4 42
Time complexity: O(n^2), The time complexity of this approach is O(n^2) as we are using memoization to store the subproblem solutions which are calculated again and again.
Auxiliary Space: O(n^2), The space complexity of this approach is O(n^2) as we are using a map of size n^2 to store the solutions of the subproblems.
Optimal Strategy for a Game using dp:
To solve the problem follow the below idea:
Since the same subproblems are called again, this problem has the Overlapping Subproblems property. So the re-computations of the same subproblems can be avoided by constructing a temporary array in a bottom-up manner using the above recursive formula.
Follow the below steps to solve the problem:
- Create a 2-D array table of size N * N
- Run a nested for loop to consider i and j at every possible position with a distance equal to ‘gap’ between them
- Declare an integer x, If (i+2) is less than or equal to j then set x equal to table[i+2][j], else equal to zero
- Declare an integer y, If (i+1) is less than or equal to j-1 then set y equal to table[i+1][j-1], else equal to zero
- Declare an integer z, If i is less than or equal to j-2 then set z equal to table[i][j-2], else equal to zero
- Set table[i][j] equal to maximum of arr[i] + min(x, y) or arr[j] + min(y, z)
- Return table[0][N-1]
Below is the implementation of the above approach:
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 code 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 code 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 contributed # 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
Time Complexity: O(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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