# 486. 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.
```

**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.

{% hint style="info" %}
dp\[i]\[j]代表第一個玩家能夠在\[i...j]區間比第二個玩家多拿多少分數。\
dp state transition: \
dp\[i]\[i] = nums\[i]\
dp\[i]\[j] = max(nums\[i] - dp\[i + 1]\[j], nums\[j] - dp\[i]\[j - 1])\
如果選擇了i那就能得到nums\[i]分，然後第二個玩家就會從\[i + 1...j]的頭尾選擇對他最有利的，所以最後用nums\[i] - dp\[i + 1]\[j]代表選擇nums\[i]的結果。如果選擇了j，那就能得到nums\[j]分，第二個玩家只能從\[i...j - 1]區間的頭尾選擇最有利的，最後用nums\[j] - dp\[i]\[j - 1]代表第一個玩家選擇nums\[j]的結果。\\

{% endhint %}

```cpp
// Dynamic Programming
bool PredictTheWinner(vector<int>& nums) { // time: O(n^2); space: O(n^2)
    int n = nums.size();
    // dp[i][j] means how much more scores player 1 can get from i to j than player 2
    vector<vector<int> > dp(n, vector<int>(n));
    for (int i = 0; i < n; ++i) dp[i][i] = nums[i];
    for (int len = 1; len < n; ++len) {
        for (int i = 0; i < n - len; ++i) {
            int j = i + len;
            dp[i][j] = max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]);
        }
    }
    return dp.front().back() >= 0;
}
```

```cpp
// Dynamic Programming
bool PredictTheWinner(vector<int>& nums) { // time: O(n^2); space: O(n^2)
    int n = nums.size();
    if (n & 1 == 0) return true;
    // dp[i][j] means how much more scores player 1 can get from i to j than player 2
    vector<int> dp(n);
    for (int i = n - 1; i >= 0; --i) {
        for (int j = i; j < n; ++j) {
            if (i == j)
                dp[j] = nums[j];
            else
                dp[j] = max(nums[i] - dp[j], nums[j] - dp[j - 1]);
        }
    }
    return dp.back() >= 0;
}
```
