arr); n + 1, 0);

// Base cases
$dp[0] = 0; // If no bricks, score is 0
`$dp[1] = $`arr[0]; // If only one brick, score is the value of that brick

// Loop to calculate maximum score
for (`$i = 2; $`i <= `$n; $`i++) {
    // At each step, we have three options: take 1, 2, or 3 bricks
    // We choose the option that minimizes the opponent's score
    `$dp[$`i] = max(
        `$arr[$`i - 1] + `$arr[$`i - 2] + (`$i >= 4 ? $`dp[$i - 3] : 0), // Take 3 bricks
        `$arr[$`i - 1] + (`$i >= 3 ? $`arr[`$i - 2] + $`dp[$i - 2] : 0), // Take 2 bricks
        `$arr[$`i - 1] + `$dp[$`i - 1] // Take 1 brick
    );
}

return `$dp[$`n]; // Return the maximum score

Here is my solution in java, javascript, python, C, C++, Csharp HackerRank Bricks Game Problem Solution

Here is Bricks Game problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-bricks-game-problem-solution.html

For anyone stuck, here's a 2 line function that shows logic of the final solution:

def makeMove(arr):
    if len(arr) <= 3: return sum(arr)
    return sum(arr) - min([makeMove(arr[i : len(arr)]) for i in range(1, 4)])

It gives correct answer, but is not optimal. In order to solve it properly, it needs to be implemented using iteration instead of recursion

Did you know that there are actually three game modes that you can play in Bricks n Balls? Yes, there are, and you can select them from the main menu where the missions are displayed. Sometimes games like these are similar to strategy games at https://wildcardcityvip.com/. But it wasn't clear to me how you embed strategies through the code.