You are given an integer array nums
. Two players are playing a game with this array: player 1 and player 2.
Player 1 and player 2 take turns, with player 1 starting first.
Both players start the game with a score of 0
.
At each turn, the player takes one of the numbers from either end of the array (i.e., nums[0]
or nums[nums.length - 1]
)
which reduces the size of the array by 1
.
The player adds the chosen number to their score.
The game ends when there are no more elements in the array.
Return true
if Player 1 can win the game.
If the scores of both players are equal, then player 1 is still the winner,
and you should also return true
. You may assume that both players are playing optimally.
Test Cases
Example 1:
Input: nums = [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: nums = [1,5,233,7]
Output: true
Explanation: Player 1 first chooses 1. Then player 2 has 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.
Solution
class Solution {
public:
bool PredictTheWinner(vector<int>& nums) {
int n = nums.size();
vector<int> dp(n, 0);
for(int start=n-1; start>=0; start--) {
for(int end=start; end<n; end++) {
if (start == end) {
dp[start] = nums[start];
} else {
int a = nums[start] - dp[end];
int b = nums[end] - dp[end - 1];
dp[end] = max(a, b);
}
}
}
return dp[n-1] >= 0;
}
};
public class Solution {
public boolean PredictTheWinner(int[] nums) {
int[] dp = new int[nums.length];
for (int start = nums.length; start >= 0; start--) {
for (int end = start; end < nums.length; end++) {
if (start == end) {
dp[start] = nums[start];
} else {
int a = nums[start] - dp[end];
int b = nums[end] - dp[end - 1];
dp[end] = Math.max(a, b);
}
}
}
return dp[nums.length - 1] >= 0;
}
}