Coins in a Line III 396

Question

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?

Solution

用memory search的方法解，和II类似。dp[i][j]表示在i－j区间里我们能取到的最大值。

我们每次能取的的值为开头或者结尾元素，因此我们在i－j区间能取的最大值为：i－j区间元素的总和－对手在剩下区间能取到的元素的总和的较小值。因此，状态函数为：

dp[i][j] = sum[i][j] - min(dp[i+1][j], dp[i][j-1])。

代码如下：

public class Solution {
    /**
     * @param values: an array of integers
     * @return: a boolean which equals to true if the first player will win
     */
    public boolean firstWillWin(int[] values) {
        // write your code here
        if(values == null || values.length == 0){
            return false;
        }

        int n = values.length;
        int[] sum = new int[n + 1];
        sum[0] = 0;
        for(int i = 1; i <= n; i++){
            sum[i] = sum[i - 1] + values[i - 1];
        }
        int[][] dp = new int[n + 1][n + 1];
        boolean[][] visit = new boolean[n + 1][n + 1];

        return search(1, n, sum, dp, visit) > sum[n] / 2;
    }

    private int search(int start, int end, int[] sum, int[][] dp, boolean[][] visit){
        if(visit[start][end]){
            return dp[start][end];
        }

        if(start == end){
            visit[start][end] = true;
            return dp[start][end] = sum[end] - sum[start - 1];
        }

        int max = (sum[end] - sum[start - 1]) - Math.min(search(start, end - 1, sum, dp, visit), search(start + 1, end, sum, dp, visit));

        visit[start][end] = true;
        dp[start][end] = max;
        return dp[start][end];
    }
}