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Last Updated: 11 Dec, 2020

# Optimal Strategy for a Game

Easy

## Problem statement

#### The rule of the game is as follows:

``````1. Each coin has a value associated with it.

2. It’s a two-player game played against an opponent with alternating turns.

3. At each turn, the player picks either the first or the last coin from the line and permanently removes it.

4. The value associated with the coin picked by the player adds up to the total amount the player wins.
``````

#### Note:

``````'N' is always even number.

Ninjax is as smart as you, so he will play so as to maximize the amount he wins.
``````
##### Example 1:
``````Let the values associated with four coins be: [9, 5, 21, 7]

Let’s say that initially, you pick 9 and Ninjax picks 7.
Then, you pick 21 and Ninjax picks 5.
So, you win a total amount of (9+21), i.e. 30.

In case you would have picked up 7 initially and Ninjax would have picked 21 (as he plays optimally).
Then, you would pick 9 and Ninjax would choose 5. So, you win a total amount of (7+9), i.e. 16, which is not the maximum you can obtain.

Thus, the maximum amount you can win is 30.
``````
##### Example 2:
``````Let the values associated with four coins be: [20, 50, 5, 10]

Let’s say that initially, you pick 10 and Ninjax picks 20.
Then, you pick 50 and Ninjax picks 5.
So, you win a total amount of (10+50), i.e. 60.

In case you would have picked up 20 initially and Ninjax would have picked 50 (as he plays optimally).
Then, you would pick 10 and Ninjax would choose 5. So, you win a total amount of (20+10), i.e. 30, which is not the maximum you can obtain.

Thus, the maximum amount you can win is 60.
``````
##### Input format:
``````The very first line of input contains an integer T denoting the number of test cases.

The first line of every test case contains an integer ‘N’ denoting the number of coins present in the line initially.

The second line of every test case contains ‘N’ space-separated integers denoting the values associated with the coins placed by Ninjax.
``````
##### Output format:
``````For each test case, print the required answer in a separate line.
``````
##### Note:
``````You do not need to print anything, it has already been taken care of. Just implement the given function.
``````
##### Constraints:
``````1 <= 'T' <= 10
2 <= 'N' <= 10 ^ 3
0 <= 'VALUE' <= 10 ^ 5

Where 'T' is the number of test cases, 'N' is the number of coins and 'VALUE' is the amount on each coin.

Time Limit: 1 sec
``````

## Approaches

### 01 Approach

• Suppose it's your turn and you are left with coins in the index range ['I', ‘J’] (other coins have already been picked up in previous turns). You have the option to pick either ith or jth coin. Of these two options, you would select the one which maximizes your winning amount.
• If you pick the ith coin. The other player will have the option to pick ('I'+1)th or ‘J’th coin.
→ If the other player picks the ('I'+1)th coin. You can pick either end of the range ['I'+2, ‘J’].
→If the other player picks ‘J’th coin. You can pick either end of the range ['I'+1, ‘J’-1].
As the other player wants to maximize his amount (thereby minimizing the amount you can win). Hence, of the two ranges which you are left with (mentioned above), you can only use that range that gives us the minimum amount.
• If you pick the jth coin. The other player will have the option to pick ith or ('J'-1)th coin.
→ If the other player picks the ith coin. You can pick either end of the range ['I'+1, ‘J’-1].
If the other player picks  ('J'-1)th coin. You can pick either end of the range ['I', ‘J’-2].
Similarly here, of the two ranges which you are left with (mentioned above), you can only use that range which gives us the minimum amount.
• We’ll use a helper function that would return the maximum amount you can win for a given sub-array of coins.

Algorithm for the above solution is as follows:

→ Int helper(vector<int> ‘COINS’, int ‘I’, int ‘J'):

• Base Case 1: if ‘I’>'J', this is not possible so return 0.
• Base Case 2: if ‘I’='J', means only one element in the sub-array, hence return coins[i].
• Select the ith coin, and recursively call the helper function for the two possible cases and select the one which gives the minimum.
• 'X' = ‘COINS’['I'] + min( helper('COINS', 'I'+1, 'J'-1), helper('COINS', 'I'+2, 'J'))
• Select the jth coin, and recursively call the helper function for the two possible cases and select the one which gives the minimum.
• ‘Y’ = ‘COINS’['I'] + min( helper('COINS', 'I'+1, 'J'-1), helper('COINS', 'I', 'J'-2))
• For the above two possible amounts, select the maximum one.
• ‘AMOUNT’ = Max('X', ‘Y’)
• Return this selected amount.

→ Int optimalStrategyOfGame(vector<int> ‘COINS’, int ‘N'):

• Call helper('COINS', 0, ‘N’ - 1)
• Return the result generated by the helper function.