486 Predict the Winner

486. Predict the Winner

1. Question

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.

2. Implementation

(1) DP

class Solution {
    public boolean PredictTheWinner(int[] nums) {
        int n = nums.length;
        // dp[i][j] means how many more scores the first player can earn in the nums[i...j] than the second player
        int[][] dp = new int[n][n];

        for (int i = 0; i < n; i++) {
            dp[i][i] = nums[i];
        }

        for (int len = 2; len <= n; len++) {
            for (int start = 0; start < n - len + 1; start++) {
                int end = start + len - 1;
                // If the first player choose the coin at the beginning, he can earn nums[start] - dp[start + 1][end] more scores
                // If the first player choose the coin at the end, he can earn nums[end] - dp[start][end - 1] more scores
                dp[start][end] = Math.max(nums[start] - dp[start + 1][end], nums[end] - dp[start][end - 1]);
            }
        }
        return dp[0][n - 1] >= 0;
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(n ^ 2), 空间复杂度O(n ^ 2)