# 32. Predict the Winner

Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available for the next player. This continues until all the scores have been chosen. The player with the maximum score wins.

Given an array of scores, predict whether player 1 is the winner. You can assume each player plays to maximize his score.

**Example 1:**

```
Input: [1, 5, 2]
Output: False
Explanation: Initially, player 1 can choose between 1 and 2. 
If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2). 
So, final score of player 1 is 1 + 2 = 3, and player 2 is 5. 
Hence, player 1 will never be the winner and you need to return False.
```

**Example 2:**

```
Input: [1, 5, 233, 7]
Output: True
Explanation: Player 1 first chooses 1. Then player 2 have to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233.
Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.
```

## Solution: (Recursion)

**Approach:**\
**Checking for both the possible cases**\
**And player 2 always plays optimally**

```cpp
class Solution
{
public:
    bool calcScore(vector<int> &nums, int s, int e, int c1, int c2, bool turn)
    {

        if (s > e)
        {
            return c1 >= c2;
        }

        if (turn)
        {
            return (calcScore(nums, s + 1, e, c1 + nums[s], c2, false) || calcScore(nums, s, e - 1, c1 + nums[e], c2, false));
        }
        else
        {
            return (calcScore(nums, s + 1, e, c1, c2 + nums[s], true) && calcScore(nums, s, e - 1, c1, c2 + nums[e], true));
        }
    }

    bool PredictTheWinner(vector<int> &nums)
    {
       return calcScore(nums, 0, nums.size() - 1, 0, 0, true);
    }
};
```

**Time Complexity: O(2^n)**

## Solution: (DP)

**Approach:**\
**Computing the maximum profit possible for each subarray**

```cpp
class Solution
{
public:
    bool PredictTheWinner(vector<int> &nums)
    {

        int n = nums.size();

        vector<vector<int>> dp(n, vector<int>(n, 0));

        for (int i = 0; i < nums.size(); i++)
        {
            dp[i][i] = nums[i];
        }

        for (int k = 2; k <= n; k++)
        {
            for (int i = 0; i <= n - k; i++)
            {
                int fs = nums[i] - dp[i + 1][i + k - 1];
                int ls = nums[i + k - 1] - dp[i][i + k - 2];
                dp[i][i + k - 1] = max(fs, ls);
            }
        }

        return dp[0][n - 1] >= 0;
    }
};
```

**Time Complexity: O(N^2)**

## Solution: (DP) **(finding value of both players)**

```cpp
// Some code
class solution
{
public:
    int maxCoins(vector<int> &arr, int n)
    {
        //Write your code here

        vector<vector<pair<int, int>>> dp(n, vector<pair<int, int>>(n));

        for (int i = 0; i < n; i++)
        {
            dp[i][i].first = arr[i];
            dp[i][i].second = 0;
        }

        for (int k = 2; k <= n; k++)
        {
            for (int i = 0; i <= n - k; i++)
            {
                int lp = arr[i] + dp[i + 1][i + k - 1].second;
                int rp = arr[i + k - 1] + dp[i][i + k - 2].second;
                if (lp > rp)
                {
                    dp[i][i + k - 1].first = lp;
                    dp[i][i + k - 1].second = dp[i + 1][i + k - 1].first;
                }
                else
                {
                    dp[i][i + k - 1].first = rp;
                    dp[i][i + k - 1].second = dp[i][i + k - 2].first;
                }
            }
        }

        return dp[0][n - 1].first;
    }
};
```


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