Coins in a Line III

Hard Dynamic Programming Array Game Theory

LintCode: http://www.lintcode.com/en/problem/coins-in-a-line-iii/

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?

Thinkin

DP:

• State:
  dp[i][j]: max sum which can get between i and j by first player
• Function:
  dp[i][j] = sum[j] - sum[i - 1] - min(dp[i][j-1], dp[i+1],[j])
• Initialize
  dp[i][i] = sum[i] - sum[i - 1]
• Answer:
  dp[0][n]

https://www.gitbook.com/book/aaronice/lintcode/details
https://aaronice.gitbooks.io/lintcode/content/dynamic_programming/coins_in_a_line_iii.html

Solution

Java

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[][] dp = new int[n + 1][n + 1];
        boolean[][] visited = new boolean[n + 1][n + 1];
        int[] sum = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            sum[i] = sum[i - 1] + values[n - i];
        }
        int max = maxValue(1, n, dp, visited, sum);
        return max > sum[n] / 2;

    }

    private int maxValue(int i, int j, int[][] dp, boolean[][] visited, int[]sum) {
        if (visited[i][j]) {
            return dp[i][j];
        }
        int subSum = sum[j] -sum[i - 1];
        if (i == j) {
            dp[i][j] = subSum;
        } else {
            dp[i][j] = subSum - Math.min(maxValue(i, j - 1, dp, visited, sum), maxValue(i + 1, j, dp, visited, sum));
        }
        visited[i][j] = true;
        return dp[i][j];
    }
}