LeetCode 877: Stone Game
LeetCode 877 Solution Explanation
Explanation:
To solve this problem, we can use dynamic programming. We can create a 2D dp array where dp[i][j] represents the maximum number of stones one player can collect from piles[i] to piles[j] if both players play optimally.
At each step, we need to consider two scenarios:
- If Alice takes piles[i], the maximum number of stones she can collect is piles[i] + dp[i+1][j] (since Bob will play optimally to maximize his stones from the remaining piles).
- If Alice takes piles[j], the maximum number of stones she can collect is piles[j] + dp[i][j-1].
Both players aim to maximize the number of stones they collect, so Alice wants to maximize her number of stones at the end, which means she wants to maximize the difference between her stones and Bob's stones. If the difference is positive, Alice wins.
After filling up the dp array, we check if dp[0][n-1] (where n is the number of piles) is greater than 0, indicating that Alice wins.
Time Complexity: O(n^2) where n is the number of piles
Space Complexity: O(n^2)
:
LeetCode 877 Solutions in Java, C++, Python
class Solution {
public boolean stoneGame(int[] piles) {
int n = piles.length;
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++) {
dp[i][i] = piles[i];
}
for (int len = 2; len <= n; len++) {
for (int i = 0; i <= n - len; i++) {
int j = i + len - 1;
dp[i][j] = Math.max(piles[i] - dp[i + 1][j], piles[j] - dp[i][j - 1]);
}
}
return dp[0][n - 1] > 0;
}
}Interactive Code Editor for LeetCode 877
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