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Stone Game II Solution Explanation

C++ Code Explanation

  1. Initialization:

    • Determine the number of piles.
    • Create a DP table dp[i][M] initialized to 0. This table will store the maximum number of stones the current player can collect starting from pile i with a given M.
    • Create a suffixSum array to store the sum of stones from the current pile to the last pile.

  2. Calculate Suffix Sums:

    • Iterate from the last pile to the first, updating suffixSum to reflect the total number of stones remaining from the current pile onward.

  3. Fill the DP Table:

    • Iterate over the piles in reverse order and over possible values of M.
    • For each pile i and value of M, iterate over the possible number of piles X the player can take (where 1 <= X <= 2 * M).
    • Update the DP table by calculating the maximum stones the current player can collect, taking into account the remaining stones and the best outcome for the opponent.

  4. Final Result:

    • The answer is the maximum stones the first player can collect starting from pile 0 with M = 1.

Java Code Explanation

  1. Initialization:

    • Determine the length of the piles array.
    • Initialize a DP table dp[i][M] to store the maximum number of stones the current player can collect starting from index i with a value of M.
    • Initialize a suffixSum array to store the sum of all elements from index i to the end.

  2. Calculate Suffix Sums:

    • Calculate suffix sums starting from the end of the array towards the beginning. The suffix sum at each index is the sum of the current pile and the suffix sum of the next pile.

  3. Fill the DP Table:

    • Iterate from the end of the array towards the beginning.
    • For each index i and value of M, iterate over the possible number of piles X that the player can take (where 1 <= X <= 2 * M).
    • Update the DP table by calculating the maximum stones the current player can collect, considering the remaining stones and the best outcome for the opponent.

  4. Final Result:

    • The answer is the maximum number of stones the first player can collect starting from index 0 with M = 1.

JavaScript Code Explanation

  1. Initialization:

    • Determine the length of the piles array.
    • Initialize a DP table dp[i][M] with dimensions (n+1) x (n+1) filled with zeros. This table stores the maximum stones the first player can collect starting from pile i with M as the maximum number of piles they can take.
    • Initialize a suffixSum array to store the sum of stones from pile i to the last pile.

  2. Calculate Suffix Sums:

    • Calculate the suffix sums from the last pile to the first. Each entry in the suffixSum array represents the total number of stones from the current pile onward.

  3. Fill the DP Table:

    • Iterate from the last pile to the first.
    • For each pile i and value of M, iterate over the possible number of piles X that the player can take (where 1 <= X <= 2 * M).
    • Update the DP table by calculating the maximum stones the current player can collect, considering the remaining stones and the best outcome for the opponent.

  4. Final Result:

    • The result is the maximum stones the first player can collect starting from the first pile with M = 1.

Python Code Explanation

  1. Initialization:

    • Determine the total number of piles.
    • Initialize a DP table dp[i][M] where each entry represents the maximum stones the current player can collect starting from pile i with a maximum of M piles to take.
    • Initialize a suffixSum array where each entry represents the total number of stones from pile i to the end.

  2. Calculate Suffix Sums:

    • Compute the suffix sums from the last pile to the first. This allows for efficient calculation of remaining stones when filling the DP table.

  3. Fill the DP Table:

    • Iterate from the last pile to the first.
    • For each pile i and value of M, iterate over the possible number of piles X that the player can take (where 1 <= X <= min(2 * M, remaining piles)).
    • Update the DP table by calculating the maximum stones the current player can collect, taking into account the remaining stones and the best outcome for the opponent.

  4. Final Result:

    • The result is the maximum stones the first player can collect starting from the first pile with M = 1.

Go Code Explanation

  1. Initialization:

    • Determine the number of piles.
    • Initialize a DP table dp[i][M] to store the maximum stones the current player can collect starting from the i-th pile with the current M value.
    • Initialize a suffixSum array to store the total number of stones from pile i to the last pile.

  2. Calculate Suffix Sums:

    • Calculate the suffix sums in reverse order. Each entry in the suffixSum array represents the total number of stones from the current pile onward.

  3. Fill the DP Table:

    • Iterate over the piles in reverse order.
    • For each pile i and value of M, iterate over the possible number of piles X that the player can take (where 1 <= X <= 2 * M).
    • Update the DP table by calculating the maximum stones the current player can collect, taking into account the remaining stones and the best outcome for the opponent.

  4. Final Result:

    • The answer is the maximum stones the first player can collect starting from the first pile with M = 1.