# Time:  O(n^2)
# Space: O(n)

# 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 <= length of the array <= 20.
# Any scores in the given array are non-negative integers and will not exceed 10,000,000.
# If the scores of both players are equal, then player 1 is still the winner.

class Solution(object):
    def PredictTheWinner(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        if len(nums) % 2 == 0 or len(nums) == 1:
            return True

        dp = [0] * len(nums);
        for i in reversed(xrange(len(nums))):
            dp[i] = nums[i]
            for j in xrange(i+1, len(nums)):
                dp[j] = max(nums[i] - dp[j], nums[j] - dp[j - 1])

        return dp[-1] >= 0
        

Create predict-the-winner.py