Formatted question description: https://leetcode.ca/all/486.html
486. Predict the Winner (Medium)
Given an array of scores that are nonnegative 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 nonnegative integers and will not exceed 10,000,000.
 If the scores of both players are equal, then player 1 is still the winner.
Companies:
Google, Works Applications
Related Topics:
Dynamic Programming, Minimax
Solution 1. DP
Let dp[len][i]
be the winning score gap of subarray starting from nums[i]
with length len
.
dp[len][i] = max{
nums[i]  dp[len  1][i],
nums[i + len  1]  dp[len  1][i]
}
Where 2 <= len <= N
and 0 <= i <= N  len
.
For len = 1
, dp[1][i] = nums[i]
.
// OJ: https://leetcode.com/problems/predictthewinner/
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
bool PredictTheWinner(vector<int>& nums) {
int N = nums.size();
unordered_map<int, unordered_map<int, int>> dp;
for (int i = 0; i < N; ++i) dp[i][1] = nums[i];
for (int len = 2; len <= N; ++len) {
for (int i = 0; i <= N  len; ++i) {
dp[i][len] = max(nums[i]  dp[i + 1][len  1], nums[i + len  1]  dp[i][len  1]);
}
}
return dp[0][N] >= 0;
}
};
Solution 2. Optimized DP
Since dp[len][i]
is only dependent on dp[len  1][i]
and dp[len  1][i]
, we can reduce the space from O(N^2)
to O(N)
.
// OJ: https://leetcode.com/problems/predictthewinner/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
bool PredictTheWinner(vector<int>& nums) {
int N = nums.size();
vector<int> dp(N);
for (int i = 0; i < N; ++i) dp[i] = nums[i];
for (int len = 2; len <= N; ++len)
for (int i = 0; i <= N  len; ++i)
dp[i] = max(nums[i]  dp[i + 1], nums[i + len  1]  dp[i]);
return dp[0] >= 0;
}
};
Java

public class Solution { public boolean PredictTheWinner(int[] nums) { if (nums == null  nums.length == 0) return false; int length = nums.length; int[][] dp = new int[length][length]; for (int i = 0; i < length; i++) dp[i][i] = nums[i]; for (int i = length  2; i >= 0; i) { for (int j = i + 1; j < length; j++) dp[i][j] = Math.max(nums[i]  dp[i + 1][j], nums[j]  dp[i][j  1]); } return dp[0][length  1] >= 0; } }

// OJ: https://leetcode.com/problems/predictthewinner/ // Time: O(N^2) // Space: O(N^2) class Solution { public: bool PredictTheWinner(vector<int>& nums) { int N = nums.size(); unordered_map<int, unordered_map<int, int>> dp; for (int i = 0; i < N; ++i) dp[i][1] = nums[i]; for (int len = 2; len <= N; ++len) { for (int i = 0; i <= N  len; ++i) { dp[i][len] = max(nums[i]  dp[i + 1][len  1], nums[i + len  1]  dp[i][len  1]); } } return dp[0][N] >= 0; } };

class Solution(object): def PredictTheWinner(self, nums): """ :type nums: List[int] :rtype: bool """ def canWin(nums, start, end, visited, partSum, maxSum, order): # print "canWin order=", order if (start, end, partSum, order) in visited: return visited[start, end, partSum, order] if start > end: # print "order=", order, partSum, maxSum if order == 0: if partSum >= maxSum  partSum: return False return True else: if partSum >= maxSum  partSum: return True return False visited[start, end, partSum, order] = False if not canWin(nums, start + 1, end, visited, partSum  order * nums[start], maxSum, ~order): visited[start, end, partSum, order] = True # print "order=", order, "return True" return True if not canWin(nums, start, end  1, visited, partSum  order * nums[end], maxSum, ~order): visited[start, end, partSum, order] = True # print "order=", order, "return True" return True return visited[start, end, partSum, order] return canWin(nums, 0, len(nums)  1, {}, 0, sum(nums), 1)