Alice and Bob continue their games with piles of stones. There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue.
Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take 1, 2 or 3 stones from the first remaining stones in the row.
The score of each player is the sum of values of the stones taken. The score of each player is 0 initially.
The objective of the game is to end with the highest score, and the winner is the player with the highest score and there could be a tie. The game continues until all the stones have been taken.
Assume Alice and Bob play optimally.
Return "Alice" if Alice will win, "Bob" if Bob will win or "Tie" if they end the game with the same score.
Example 1:
Input: values = [1,2,3,7]
Output: "Bob"
Explanation: Alice will always lose. Her best move will be to take three piles and the score become 6. Now the score of Bob is 7 and Bob wins.
Example 2:
Input: values = [1,2,3,-9]
Output: "Alice"
Explanation: Alice must choose all the three piles at the first move to win and leave Bob with negative score.
If Alice chooses one pile her score will be 1 and the next move Bob's score becomes 5. The next move Alice will take the pile with value = -9 and lose.
If Alice chooses two piles her score will be 3 and the next move Bob's score becomes 3. The next move Alice will take the pile with value = -9 and also lose.
Remember that both play optimally so here Alice will choose the scenario that makes her win.
Example 3:
Input: values = [1,2,3,6]
Output: "Tie"
Explanation: Alice cannot win this game. She can end the game in a draw if she decided to choose all the first three piles, otherwise she will lose.
Example 4:
Input: values = [1,2,3,-1,-2,-3,7]
Output: "Alice"
Example 5:
Input: values = [-1,-2,-3]
Output: "Tie"
Constraints:
1 <= values.length <= 50000
-1000 <= values[i] <= 1000
代码
Approach #1 DP
Time: O(1) && Space: O(1)
Let dp[i] denote the maximum score one can get if he/she takes stones first at the position i. We only need to compare dp[0]*2 with the total sum of stone values to determine the game result.
AlgorithmStep1. Calculate the suffix sum of stone values. Step2. For i = n-1, n-2, …, 0, calculate dp[i]. We need to enumerate three possible scenarios which correspond to taking 1, 2, 3 stones at this round. state transition: dp[i] = max(dp[i], suffixSum[i]-suffixSum[k+1] + suffixSum[k+1] - dp[k+1]) = max(dp[i], suffixSum[i] - dp[k+1])，for k = i, i+1, i+2, where (suffixSum[k+1] - dp[k+1]) means the score one can get when he/she takes stones secondly at the position k+1.Step3. Compare suffixSum[0] with dp[0]*2. (Alice score: dp[0], Bob score: suffixSum[0]-dp[0])
Complexity Time: O(n) Space: O(n) In fact, we could reduce the space complexity to O(1) since we only need suffixSum[i:i+3] and dp[i:i+3] in each iteration of Step2. To make the code clear to understand, the space optimization is not implemented in the code below.
class Solution {
public String stoneGameIII(int[] stoneValue) {
int n = stoneValue.length;
int[] suffixSum = new int[n+1];
int[] dp = new int[n+1];
suffixSum[n] = 0;
dp[n] = 0;
for (int i = n - 1; i >= 0; i--) {
suffixSum[i] = stoneValue[i] + suffixSum[i+1];
}
for (int i = n - 1; i >= 0; i--) {
dp[i] = stoneValue[i] + suffixSum[i+1] - dp[i+1];
for (int k = i + 1; k < i + 3; && k < n; k++) {
dp[i] = Math.max(dp[i], suffixSum[i] - dp[k+1]);
}
}
/** alternative
Arrays.fill(dp, Integer.MIN_VALUE);
for (int i = n-1; i >= 0; i--) {
for (int k = i; k < i+3 && k < n; k++) {
dp[i] = Math.max(dp[i], (suffixSum[i]-suffixSum[k+1]) + (suffixSum[k+1]-dp[k+1]));
}
}
**/
if (dp[0] * 2 == suffixSum[0]) {
return "Tie";
} else if (dp[0] * 2 > suffixSum[0]) {
return "Alice";
} else {
return "Bob";
}
}
}