# Predict the Winner

## Description

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:**<br>

```
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.
```

**Note:**

1. 1 <= length of the array <= 20.
2. Any scores in the given array are non-negative integers and will not exceed 10,000,000.
3. If the scores of both players are equal, then player 1 is still the winner.

## Solutions

### My solution

This problem is similar to the problem [Stone Game](https://twchen.gitbook.io/leetcode/dynamic-programming/stone-game). The difference is that the number of numbers can be odd in this problem.

Let n be the number of numbers.

If n is even, by the analysis in [Stone Game](https://twchen.gitbook.io/leetcode/dynamic-programming/stone-game), player 1 always wins by choosing either all even-indexed numbers or all odd-indexed numbers.

If n is odd, let `dp[i, j]` be the maximum score that player 1 can get if the given numbers are `nums[i..j]`.

If player 1 picks `nums[i]`, the score that player 1 will get is `nums[i] + (sums[i+1..j] - dp[i+1, j]) = sums[i..j] - dp[i+1, j]`.

Similarly, if player 1 picks `nums[j]`, the score that player 1 will get is `sums[i..j] - dp[i..j-1]`.

Therefore, `dp[i, j] = sums[i..j] - min(dp[i+1, j], dp[i, j-1])`.

```cpp
class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int n = nums.size();
        if(n % 2 == 0) return true;
        // dp[i][j] = the maximum score that the first player can get given the nums are nums[i..j]
        vector<vector<int>> dp(n, vector<int>(n));
        // sums[i] = sum of the first i numbers
        vector<int> sums(n + 1);
        for(int i = 0; i < n; ++i){
            sums[i + 1] = sums[i] + nums[i];
            dp[i][i] = nums[i];
        }
        for(int step = 1; step < n; ++step){
            for(int i = 0; i + step < n; ++i){
                int j = i + step;
                dp[i][j] = sums[j + 1] - sums[i] - min(dp[i + 1][j], dp[i][j - 1]);
            }
        }
        return 2 * dp[0][n - 1] >= sums[n];
    }
}; 
```

### A better solution

Define `dp[i, j]` to be how much more scores that player 1 can get than player 2, given the numbers are `nums[i..j]`.

Then `dp[i, j] = max(nums[i] - dp[i+1, j], nums[j] - dp[i, j-1])`. So we don't need the sum of `nums[i..j]`.

```cpp
class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int n = nums.size();
        if(n % 2 == 0) return true;
        vector<vector<int>> dp(n, vector<int>(n));
        for(int i = 0; i < n; ++i){
            dp[i][i] = nums[i];
        }
        for(int step = 1; step < n; ++step){
            for(int i = 0; i + step < n; ++i){
                int j = i + step;
                dp[i][j] = max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]);
            }
        }
        return dp[0][n - 1] >= 0;
    }
}; 
```


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