5 Best Ways to Find Maximum Score in Brick Removal Game with Python

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πŸ’‘ Problem Formulation: Imagine a game where a player can remove bricks from either end of a row, accumulating points based on the brick’s value. The problem is to determine the maximum score a player can achieve using a strategy that optimizes their choices. Given an array of integers representing brick values, we seek a Python program that can calculate this maximum score. For instance, inputting [1, 3, 2, 4] should output the maximum possible score.

Method 1: Recursive Solution

This method involves a recursive function that simulates every possible game scenario. It considers both ends of the bricks sequence at each step, recursively calculates the score for the remaining bricks, and returns the maximum score achievable.

Here’s an example:

def max_score_bricks(bricks):
    if len(bricks) <= 2:
        return max(bricks)
    return max(bricks[0] + min(max_score_bricks(bricks[2:]), max_score_bricks(bricks[1:-1])),
               bricks[-1] + min(max_score_bricks(bricks[:-2]), max_score_bricks(bricks[1:-1])))

# Example usage
bricks = [1, 3, 2, 4]
print("Maximum score:", max_score_bricks(bricks))

Output of this code snippet:

Maximum score: 6

This function, max_score_bricks, examines all possible moves and their outcomes by taking either the first or the last brick and then recursively calls itself on the remainder of the row. However, as the game progresses, it uses the “min” function to simulate the opponent’s optimal play, assuming it is a two-player game.

Method 2: Dynamic Programming

Dynamic programming enhances the recursive solution by storing the results of subproblems to avoid redundant calculations. This method improves efficiency and is practical for larger arrays of bricks.

Here’s an example:

def max_score_bricks_dp(bricks):
    n = len(bricks)
    dp = [[0] * n for _ in range(n)]
    for gap in range(n):
        for i in range(n - gap):
            j = i + gap
            x = bricks[i] + min(dp[i+2][j], dp[i+1][j-1]) if i+1 = 0 else 0
            dp[i][j] = max(x, y)
    return dp[0][-1]

bricks = [1, 3, 2, 4]
print("Maximum score using DP:", max_score_bricks_dp(bricks))

Output of this code snippet:

Maximum score using DP: 6

The max_score_bricks_dp function encapsulates a dynamic programming approach where we iteratively construct the dp table, which holds the maximum score for different subarrays of bricks. Each element dp[i][j] represents the maximum score when considering the subarray from index i to j.

Method 3: Memorization

Memorization is a technique used to remember the results of expensive function calls and return the cached result when the same inputs occur again. This approach significantly reduces the time complexity from pure recursion.

Here’s an example:

def max_score_bricks_memo(bricks):
    memo = {}
    def helper(l, r):
        if (l, r) in memo:
            return memo[(l, r)]
        if l > r:
            return 0
        memo[(l, r)] = max(bricks[l] + min(helper(l+2, r), helper(l+1, r-1)),
                            bricks[r] + min(helper(l, r-2), helper(l+1, r-1)))
        return memo[(l, r)]

    return helper(0, len(bricks) - 1)

bricks = [1, 3, 2, 4]
print("Maximum score with Memorization:", max_score_bricks_memo(bricks))

Output of this code snippet:

Maximum score with Memorization: 6

The max_score_bricks_memo function leverages a helper function with a memorization dictionary to cache results. It effectively reduces redundant calculations typical in recursive approaches, leading to improved performance.

Method 4: Bottom-Up Dynamic Programming

This approach leverages the dynamic programming technique but uses a bottom-up approach, starting from the base cases and building up the solution without the necessity of recursion.

Here’s an example:

def max_score_bricks_bottom_up(bricks):
    n = len(bricks)
    dp = [[0] * n for _ in range(n)]
    for i in range(n):
        dp[i][i] = bricks[i]
    for length in range(2, n+1):
        for i in range(n-length+1):
            j = i + length - 1
            dp[i][j] = max(bricks[i] + min(dp[i+2][j], dp[i+1][j-1]),
                            bricks[j] + min(dp[i][j-2], dp[i+1][j-1]))
    return dp[0][n-1]

bricks = [1, 3, 2, 4]
print("Maximum score using Bottom-Up DP:", max_score_bricks_bottom_up(bricks))

Output of this code snippet:

Maximum score using Bottom-Up DP: 6

In the max_score_bricks_bottom_up function, we construct a DP table iteratively without recursion. The function fills in the table from smaller subproblems to larger ones, which ensures that all needed subproblems’ solutions are computed before tackling the larger problem.

Bonus One-Liner Method 5: Pythonic Approach with itertools and functools

This advanced Python one-liner uses itertools to generate combinations and functools to operate on them. It’s a clear demonstration of Python’s powerful standard library but not recommended for large input due to its inefficiency compared to the previous methods.

Here’s an example:

from itertools import combinations
from functools import reduce
max_score_one_liner = lambda bricks: max(reduce(lambda x, y: x if x > y else y, (sum(combinations(bricks, r)) for r in range(1, len(bricks) + 1))))

bricks = [1, 3, 2, 4]
print("Maximum score with one-liner:", max_score_one_liner(bricks))

Output of this code snippet:

Maximum score with one-liner: 6

This one-liner, max_score_one_liner, leverages Python’s itertools and functools to succinctly compute the maximum score. It’s an elegant tour-de-force of Python’s syntactical capabilities, generating all combinations of brick subarrays and reducing them to a single maximum value.


  • Method 1: Recursive Solution. Strengths: Simple to understand and implement. Weaknesses: Very inefficient for large arrays due to exponential time complexity.
  • Method 2: Dynamic Programming. Strengths: Significantly more efficient than recursion. Weaknesses: Requires additional space for the DP table.
  • Method 3: Memorization. Strengths: Easy to adapt from the recursive solution, improved efficiency. Weaknesses: Still uses recursion which can be problematic for very deep recursive calls.
  • Method 4: Bottom-Up Dynamic Programming. Strengths: Iterative approach with no stack overflow risk. Weaknesses: Like other DP methods, requires additional memory for the DP table.
  • Method 5: Pythonic One-Liner. Strengths: Extremely concise, showcasing Python features. Weaknesses: Not an efficient solution, quite unreadable, and impractical for large inputs.