LeetCode 1406: Stone Game III Solution
Master LeetCode problem 1406 (Stone Game III), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1406. Stone Game III
Problem Explanation
Explanation:
To solve this problem, we can use dynamic programming with memoization. We will create a recursive function that represents the maximum score difference that can be achieved starting from a certain index in the array. The base cases will be when we reach the end of the array, and the recursive cases will involve choosing 1, 2, or 3 stones at a time and considering the best strategy for the opponent.
We will maintain an array to keep track of the maximum score difference that can be achieved starting from each index. At each step, we will consider all possible moves and choose the one that maximizes the score difference.
After calculating the maximum score difference for Alice, Bob, and tie scenarios, we can determine the winner based on these values.
Time Complexity:
The time complexity of this approach is O(n), where n is the number of stones in the array.
Space Complexity:
The space complexity is also O(n) due to the memoization array.
:
Solution Code
class Solution {
public String stoneGameIII(int[] stoneValue) {
int n = stoneValue.length;
int[] dp = new int[n];
Arrays.fill(dp, Integer.MIN_VALUE);
int scoreDiff = getMaxScoreDiff(0, stoneValue, dp);
if (scoreDiff > 0) {
return "Alice";
} else if (scoreDiff < 0) {
return "Bob";
} else {
return "Tie";
}
}
private int getMaxScoreDiff(int index, int[] stoneValue, int[] dp) {
if (index >= stoneValue.length) {
return 0;
}
if (dp[index] != Integer.MIN_VALUE) {
return dp[index];
}
int maxDiff = Integer.MIN_VALUE;
int curSum = 0;
for (int i = 0; i < 3 && index + i < stoneValue.length; i++) {
curSum += stoneValue[index + i];
maxDiff = Math.max(maxDiff, curSum - getMaxScoreDiff(index + i + 1, stoneValue, dp));
}
dp[index] = maxDiff;
return maxDiff;
}
}Try It Yourself
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