LeetCode 2029: Stone Game IX Solution

Problem Explanation

Explanation:

To solve this problem, we can use the concept of dynamic programming. We can keep track of the cumulative sum of the stones and the remainder when dividing the sum by 3. Based on these remainders, we can determine the winning strategy for Alice and Bob.

We initialize an array count to keep track of the count of stones with remainder 0, 1, and 2 when divided by 3.
We iterate over the stones and count the number of stones with remainder 0, 1, and 2 when divided by 3.
We consider three cases:
1. If the count of stones with remainder 0, 1, and 2 are all zero, it means Bob wins automatically.
2. If the counts of stones with remainder 1 and 2 are equal, Alice can remove stones with these remainders alternatively to keep the total sum not divisible by 3, and she wins.
3. If the counts of stones with remainder 1 and 2 differ by more than 2, Alice can choose to remove stones with the more frequent remainder to win.

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Solution Code

class Solution {
    public boolean stoneGameIX(int[] stones) {
        int[] count = new int[3];
        
        for (int stone : stones) {
            count[stone % 3]++;
        }
        
        if (count[0] % 2 == 0) {
            return count[1] != 0 && count[2] != 0;
        }
        
        return Math.abs(count[1] - count[2]) > 2 || count[0] == 0;
    }
}

Frequently Asked Questions

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2029. Stone Game IX