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.
Approach #1: DP. [Java]
class Solution {
public boolean PredictTheWinner1(int[] nums) {
int n = nums.length;
int[][] dp = new int[n+1][n+1];
for (int i = 0; i < n; ++i)
dp[i][i] = nums[i];
for (int len = 1; len <= n; ++len) {
for (int i = 0; i < n - len; ++i) {
int j = i + len;
dp[i][j] = Math.max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]);
}
}
return dp[0][n-1] >= 0;
}
public boolean PredictTheWinner(int[] nums) {
int n = nums.length;
int[] dp = new int[n+1];
for (int i = n-1; i >= 0; --i) {
for (int j = i; j < n; ++j) {
if (i == j) {
dp[i] = nums[i];
} else {
dp[j] = Math.max(nums[i] - dp[j], nums[j] - dp[j-1]);
}
}
}
return dp[n-1] >= 0;
}
}
Approach:
Using a two dimensional array dp[i][j] to save all the intermediate states. From step 1, we may see at least two ways of doing it. It just turned out that if we choose to save how much more score that the first-in-action player will earn from position i to j in the array, the code will be better in a couple of ways.
After we decide that dp[i][j] saves how much more scores that the first-in-action player will get from i to j than the second player, the next step is how we update the dp table from one states to the next. Going back to the question, each player can pick one number either from the left or the right end of the array. Suppose they are picking up numbers from position i to j in the array and it is player A's turn to pick the number now. If player A picks position i, player A will earn nums[i] score instantly. Then player B will choose from i + 1 to j. Please note that dp[i+1][j] already saves how much score that the first-in-action player will get from i + 1 to j than the second player. So it means that player B will eventually earn dp[i+1][j] more score from i+1 to j than player A. So if player A picks position i, eventually player A will get nums[i] - dp[i+1][j] more score than player B after they pick up all numbers. Similarly, if player A picks position j, player A will earn nums[i] - dp[i][j-1] more score than player B after they pick up all numbers. Since A is smart, A will always choose the max in those two options, so: dp[i][j] = Math.max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]);
Now we have the recursive formula, the next step is to decide where it all starts. This step is easy because we can easily recognize that we can start from dp[i][i], where dp[i][i] = nums[i]. Then the process becomes a very commonly seen process to update the dp table. I promise that this is a very useful process. Everyone who is preparing for interviews should get comfortable with this process:
Using a 5*5 dp table as an example, where i is the row number and j is the column number. Each dp[i][j] corresponds to a block at row i, column j on the table. We may start from filling dp[i][j], which are all the diagonal blocks. I marked them as 1. Then we can see that each dp[i][j] dpends only on dp[i+1][j] and dp[i][j-1]. On the table, it means each block (i, j) only depends on the block to its left (i, j - 1) and to its down (i + 1, j). So after filling all the blocks markes as 1, we can start to calculate those blocks marked as 2. After that, all blocks marked as 3 and so on.
| 1 | 2 | 3 | 4 | 5 |
| 1 | 2 | 3 | 4 | |
| 1 | 2 | 3 | ||
| 1 | 4 | |||
| 1 |
So in my code, I always use len to denote how far the block is away from the diagonal. So len ranges from 1 to n-1. Remeber this is the outer loop. The inner loop is all valid i positions. After filling all the upper side of the table, we will get our answer at dp[0][n-1] (marked as 5). This is the end of my code.
However, we can improve our code in the aspect of space complexity. So far, we are using a n*n matrix so the space complexity if O(n^2). It actually can be improved to O(n). That can be done by changing our way of filling the table. We may use only one dimensional dp[i] and we start to fill the table at the bottom right corner where dp[4] = nums[4]. On the next step, we start to fill the second to the last line, where it starts from dp[3] = nums[3]. Then dp[4] = Math.max(nums[4] - dp[3], nums[3] - dp[4]).