[leetcode] predict the winner

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.

A perfect discuss for this solution could be found here.
Assume the length of input is $N$.
1. if $N$ is even.
suppose the input is: $V_1, V_2, V_3, V_4, ... , V_n$.
compare the odd sequence sum with even sequence sum, choose the greater set(you could always do so)
example: [1, 5, 233, 7], odd = 234, even = 12. Choose 1 first, no matter 5 or 7 the opponent is choosing, you could always finish choosing 233 and win.
2. if $N$ is odd.
define $v(i, j)$ as max profit if it is our turn and only with coin $V_i, V_{i + 1},...,V_j$ .

$$v(i,j) = max(V_i + <range\;becomes(i + 1, j)>, V_j + <range\;becomes(i, j - 1)>)$$
pay attention here: and are all opponent’s turn to pick at his/her best to win. The assumption for this game is the opponent also make optimized moves.
Opponent make best moves => gain most profit => ensure we gain less profit
Thus:
$$v(i,j) = max(V_i + min(V(i + 2, j), V(i+1,j-1)), V_j + min(V(i, j-2), V(i+1,j-1)))$$

Code:

class Solution 
{
    vector<vector<int>> dp;
public:
    bool PredictTheWinner(vector<int>& nums)
    {
        int N = nums.size();
        if (N % 2 == 0) return true;
        dp.resize(N, vector<int>(N, -1));
        int maxprofit = helper(nums, 0, N - 1);
        return maxprofit * 2 >= accumulate(nums.begin(), nums.end(), 0);
    }
    int helper(vector<int> &nums, int i, int j)
    {
        if (i > j) return 0;
        if (dp[i][j] != -1) return dp[i][j];
        int choose_i = nums[i] + min(helper(nums, i + 1, j - 1), helper(nums, i + 2, j));
        int choose_j = nums[j] + min(helper(nums, i + 1, j - 1), helper(nums, i, j - 2));
        return dp[i][j] = max(choose_i, choose_j);
    }
};