Most Detailed Explanation || Beats 100% || C++
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to analyze the counts of stones with values congru
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to analyze the counts of stones with values congruent to 0, 1, and 2 modulo 3. By understanding the relationship between these counts, we can determine if the first player can win the game or not.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Start by counting the number of stones with values congruent to 0, 1, and 2 modulo 3. We can use an array count to keep track of these counts.\n\n- Check if the count of stones with value congruent to 0 modulo 3 is even. If it is, the first player cannot lose because they can always select a stone with value congruent to 0 modulo 3 and force the second player to make a move.\n\n- If the count of stones with value congruent to 0 modulo 3 is odd, we need to check if there is a significant difference between the counts of stones with values congruent to 1 and 2 modulo 3. If the absolute difference is greater than 2, the first player can strategically select stones to force the second player into making a losing move.\n\n- Return the appropriate result based on the conditions mentioned above. If the first player can win, return true; otherwise, return false.\n\n- By analyzing the counts of stones modulo 3 and considering the parity of the count of stones with value congruent to 0 modulo 3, we can determine if the first player has a winning strategy. This approach avoids the need for simulating the game and directly uses the properties of the stone values to make the decision.\n\n# **Explaination**\nThis solution is based on the observation that the outcome of the game only depends on the counts of stones with values congruent to 0, 1, and 2 modulo 3.\n\n- The code uses an array count of size 3 to keep track of the counts. Initially, all elements of count are set to 0.\n\n- The for loop iterates over each stone value in the stones vector. For each stone, it increments the count corresponding to its value modulo 3. This is done using the expression ++count[stone % 3].\n\nAfter counting the stones, the code checks the conditions for winning the game:\n\n- If the count of stones with value congruent to 0 modulo 3 is even (count[0] % 2 == 0), it means the first player can take all the stones with this value, and the second player can do the same. In this case, the first player cannot lose since they have an even number of stones to choose from. However, if either the count of stones with value congruent to 1 modulo 3 or the count of stones with value congruent to 2 modulo 3 is 0 (count[1] == 0 or count[2] == 0), then the first player cannot win since they have no stones to choose from that give them a valid move. Therefore, the condition count[1] > 0 && count[2] > 0 is checked.\n\n- If the count of stones with value congruent to 0 modulo 3 is odd (count[0] % 2 == 1), it means the first player can take one stone with this value, and the second player can do the same. In this case, the first player can potentially lose if there is a significant difference in the counts of stones with values congruent to 1 modulo 3 and 2 modulo 3. To check for this condition, the absolute difference between count[1] and count[2] is calculated using std::abs(count[1] - count[2]), and it is checked if it is greater than 2 (std::abs(count[1] - count[2]) > 2).\n\nBased on these conditions, the code returns true if the first player wins and false otherwise.\n\n# Code\n```\nconst static auto initialize = [] {\n    // Function to initialize input/output operations for faster execution\n    std::ios::sync_with_stdio(false); // Allows faster input/output\n    std::cin.tie(nullptr); // Ensures input is synchronized with output\n    std::cout.tie(nullptr); // Ensures output is synchronized with input\n    return nullptr;\n}();\n\nclass Solution {\npublic:\n    bool stoneGameIX(vector<int>& stones) {\n        // Create an array to count the different types of stones\n        auto count = array<int, 3>{};\n        \n        // Count the different types of stones\n        for (auto stone : stones) {\n            ++count[stone % 3];\n        }\n        \n        // Check if the first player can win the game based on the stone counts\n        return count[0] % 2 == 0 ?\n            count[1] > 0 && count[2] > 0 :\n            abs(count[1] - count[2]) > 2;\n    }\n};\n\n```
['C++']