Stone Game II

The problem is a game scenario played by Alice and Bob. The game is played with a number of piles arranged in a row, with each pile containing a positive number of stones. The goal of the game is to end with the most stones, and the problem asks to return the maximum number of stones Alice can get if both players play optimally.

We can categorize this problem as follows:

  • Domain: Game Theory
  • What Components:
    • Stone Piles: A sequence of stone piles arranged in a row, represented by an array of integers.
    • Turn-based Game: Alice and Bob take turns playing the game, following certain rules defined in the problem statement.
    • Dynamic Programming (DP): A typical approach to solving such problems where the solution can be found by solving subproblems.

Explanation and Categorizations

  1. Stone Piles: The stone piles form the basis of the game, and the players’ actions are based on the current state of these piles. Understanding the structure of these piles is essential for constructing a solution.

  2. Turn-based Strategy: The game’s rules define the strategies for each player, and the solution must consider how each player would play optimally. In this case, the constraint that each player can take stones from the first X remaining piles, where (1 \leq X \leq 2M), and then set (M = \max(M, X)), adds complexity to the game’s strategy.

  3. Dynamic Programming Solution: Given the nature of the problem, where the outcome depends on the previous actions, a dynamic programming approach can be a suitable way to find the optimal solution. By using memoization or a bottom-up approach, we can avoid redundant computations.

Solution

Here’s a Python function that implements a dynamic programming solution for this problem:

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class Solution:
    def stoneGameII(self, piles: List[int]) -> int:
        n = len(piles)
        prefix_sum = [0] * (n + 1)
        for i in range(n - 1, -1, -1):
            prefix_sum[i] = prefix_sum[i + 1] + piles[i]

        memo = {}

        def dp(index, M):
            if (index, M) in memo:
                return memo[(index, M)]

            if index + 2 * M >= n:
                return prefix_sum[index]

            best = 0
            for X in range(1, 2 * M + 1):
                next_M = max(M, X)
                best = max(best, prefix_sum[index] - dp(index + X, next_M))

            memo[(index, M)] = best
            return best

        return dp(0, 1)

The solution calculates a prefix sum of the piles in reverse, making it easier to calculate the sum of any remaining piles. Then it uses a recursive function with memoization to calculate the best score for Alice at any given state, defined by the current index and value of (M). By iterating over all possible values of (X), we can find the optimal solution.

10 Prerequisite LeetCode Problems

“1140. Stone Game II” belongs to the category of dynamic programming problems. It requires understanding of recursion, memoization, and game theory concepts. Here are 10 problems to build the required foundation:

  1. Climbing Stairs (LeetCode 70): This is a basic dynamic programming problem that can help build your understanding of DP concepts.

  2. Best Time to Buy and Sell Stock (LeetCode 121): This problem involves finding the optimal time to buy and sell stocks, a simpler problem that can help with understanding how to solve optimization problems.

  3. House Robber (LeetCode 198): This problem shares similarities with “Stone Game II” as it involves choosing maximum value with restrictions.

  4. Min Cost Climbing Stairs (LeetCode 746): This is a dynamic programming problem that can help you understand how to define states and transitions.

  5. Coin Change (LeetCode 322): This problem is about finding the fewest number of coins that you need to make up a given amount, which is a classic DP problem.

  6. Longest Increasing Subsequence (LeetCode 300): This problem can help with understanding the concept of optimal substructure in DP problems.

  7. Stone Game (LeetCode 877): It is very similar to “Stone Game II”, but simpler. This problem can help build the initial understanding needed for “Stone Game II”.

  8. Counting Bits (LeetCode 338): This problem is a simpler dynamic programming problem that can help reinforce your understanding of DP concepts.

  9. Partition Equal Subset Sum (LeetCode 416): This problem involves dividing a set into two with equal sums, a problem that helps in understanding partitioning problems.

  10. Longest Palindromic Subsequence (LeetCode 516): This problem, which involves finding the longest palindromic subsequence in a string, can help understand the concept of subsequence problems in dynamic programming.

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class Solution:
    def helper(self, piles, dp, suffixSum, i, M):
        if i == len(piles):
            return 0
        if i + 2 * M >= len(piles):
            return suffixSum[i]

        if dp[i][M] != 0:
            return dp[i][M]

        result = 0
        for x in range(1, 2 * M + 1):
            result = max(result, suffixSum[i] - self.helper(piles, dp, suffixSum, i + x, max(M, x)))

        dp[i][M] = result
        return result

    def stoneGameII(self, piles):
        if not piles:
            return 0
        dp = [[0] * len(piles) for _ in range(len(piles))]

        suffixSum = [0] * len(piles)
        suffixSum[-1] = piles[-1]
        for i in range(len(piles) - 2, -1, -1):
            suffixSum[i] = piles[i] + suffixSum[i + 1]

        return self.helper(piles, dp, suffixSum, 0, 1)

Problem Classification

Alice and Bob continue their games with piles of stones. There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones. Alice and Bob take turns, with Alice starting first. Initially, M = 1. On each player’s turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M. Then, we set M = max(M, X). The game continues until all the stones have been taken. Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.

Example 1:

Input: piles = [2,7,9,4,4] Output: 10 Explanation: If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get 2 + 4 + 4 = 10 piles in total. If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get 2 + 7 = 9 piles in total. So we return 10 since it’s larger.

Example 2:

Input: piles = [1,2,3,4,5,100] Output: 104

Constraints:

1 <= piles.length <= 100 1 <= piles[i] <= 10^4

Language Agnostic Coding Drills

This code solves the problem of the Stone Game II, where players take turns to take stones from piles. It uses several important concepts, which could be learned and practiced separately before putting it all together. Here is a possible breakdown:

  1. Defining and Calling Functions: In most modern languages, functions are a key building block of any application. You’ll need to understand how to define your own functions and how to call them.

  2. Recursion: The code uses a recursive function helper. Understanding how recursion works is crucial to understanding this code. This is a more complex concept than the previous ones and takes some practice to master.

  3. Using Arrays: The code uses several arrays (or lists in Python), so you need to understand how to create arrays, how to access their elements, and how to update them.

  4. Using Loops: The code has a for-loop to iterate over the array elements. You need to understand how loops work in general, and how for-loops work specifically.

  5. Using Conditional Statements: If-statements are used to decide what code to execute. Understanding how they work is crucial to this code.

  6. Dynamic Programming: The code uses a dynamic programming (DP) approach to solve the problem. DP is an advanced topic and understanding it can take some time and practice.

  7. Working with 2D Arrays: The code uses a 2D array (a list of lists in Python) to store intermediate results for the DP solution. Understanding how to create, access and update 2D arrays is important for this code.

  8. Suffix Sum: The code precomputes a suffix sum array, which is a common technique in competitive programming. It is crucial to understand this concept.

  9. Maximization: The problem is about maximizing the number of stones, which involves keeping track of a running maximum.

  10. Problem Specific Logic: Finally, understanding the specific rules of the Stone Game II and how they are implemented in the code is crucial. For example, why are the function arguments i and M updated the way they are, and how does the function decide when to return the sum of remaining piles.

Targeted Drills in Python

1. Defining and Calling Functions:

Drill: Write a function that adds two numbers and returns the result.

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def add_numbers(a, b):
    return a + b

print(add_numbers(3, 4))  # It should print: 7

2. Recursion:

Drill: Write a recursive function to calculate the factorial of a number.

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def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n-1)

print(factorial(5))  # It should print: 120

3. Using Arrays:

Drill: Create an array with numbers 1 to 5, print the 3rd element and update it to 10.

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numbers = [1, 2, 3, 4, 5]
print(numbers[2])  # It should print: 3
numbers[2] = 10
print(numbers)     # It should print: [1, 2, 10, 4, 5]

4. Using Loops:

Drill: Print all numbers in an array.

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numbers = [1, 2, 3, 4, 5]
for num in numbers:
    print(num)

5. Using Conditional Statements:

Drill: Write a function that takes a number and returns whether it’s positive or negative.

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def check_number(num):
    if num >= 0:
        return 'Positive'
    else:
        return 'Negative'

print(check_number(5))    # It should print: Positive
print(check_number(-3))   # It should print: Negative

6. Dynamic Programming:

Drill: Write a DP solution to calculate the nth Fibonacci number.

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def fib(n):
    dp = [0, 1] + [0] * (n-1)
    for i in range(2, n+1):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n]

print(fib(10))  # It should print: 55

7. Working with 2D Arrays:

Drill: Create a 2D array with two rows and three columns, and print the element in the second row and third column.

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arr = [[1, 2, 3], [4, 5, 6]]
print(arr[1][2])  # It should print: 6

8. Suffix Sum:

Drill: Given an array of numbers, create a new array where each element is the sum of all elements to its right in the original array.

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def suffix_sum(arr):
    result = [0] * len(arr)
    result[-1] = arr[-1]
    for i in range(len(arr) - 2, -1, -1):
        result[i] = arr[i] + result[i+1]
    return result

print(suffix_sum([1, 2, 3, 4, 5]))  # It should print: [15, 14, 12, 9, 5]

9. Maximization:

Drill: Given an array, find and print the maximum element.

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numbers = [1, 2, 10, 4, 5]
maximum = numbers[0]
for num in numbers:
    maximum = max(maximum, num)
print(maximum)  # It should print: 

10

10. Problem Specific Logic:

Drill: Given a list of piles of stones, find the maximum number of stones you can take if you can take 1, 2 or 3 piles at a time.

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def max_stones(piles):
    n = len(piles)
    dp = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        dp[i] = max(piles[i] - dp[i+1], 
                    piles[i] + piles[i+1] - dp[i+2] if i+1 < n else 0, 
                    piles[i] + piles[i+1] + piles[i+2] - dp[i+3] if i+2 < n else 0)
    return dp[0] if dp[0] > 0 else 0

print(max_stones([5, 3, 4, 5]))  # It should print: 9