[LintCode] Coins in a Line III

Coins in a Line III

There are n coins in a line. Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.

Could you please decide the first player will win or lose?

Example

Given array A = [3,2,2], return true.

Given array A = [1,2,4], return true.

Given array A = [1,20,4], return false.

Challenge

Follow Up Question:

If n is even. Is there any hacky algorithm that can decide whether first player will win or lose in O(1) memory and O(n) time?

 

備忘錄，dp[left][right]表示從left到right所能取到的最大值，因為雙方都取最優策略，所以取完一個后，對手有兩種選擇，我們要加讓對手得到更多value的那種方案，也就是自己得到更少value的方案。

class Solution {
public:
    /**
     * @param values: a vector of integers
     * @return: a boolean which equals to true if the first player will win
     */
    bool firstWillWin(vector<int> &values) {
        // write your code here
        int n = values.size();
        vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
        int sum = 0;
        for (auto v : values) sum += v;
        return sum < 2 * dfs(dp, values, 0, n - 1);
    }
    int dfs(vector<vector<int>> &dp, vector<int> &values, int left, int right) {
        if (dp[left][right] != -1) return dp[left][right];
        if (left == right) {
            dp[left][right] = values[left];
        } else if (left > right) {
            dp[left][right] = 0;
        } else {
            int take_left = min(dfs(dp, values, left + 2, right), dfs(dp, values, left + 1, right - 1)) + values[left];
            int take_right = min(dfs(dp, values, left, right - 2), dfs(dp, values, left + 1, right - 1)) + values[right];
            dp[left][right] = max(take_left, take_right);
        }
        return dp[left][right];
    }
};