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You are given an array of integers coins where coins[i] is the value of the i-th coin. There are two players playing a game optimally, alternate turns, where the first player starts picking either the first coin or the last coin from the row. The player who collects the coins with the maximum total value is the winner.

Return the winning player. If the total coins collected by the first player is more than the second player, return 1. Otherwise, if the total coins collected by second player is more than the first player, return 2. If both players collect the same amount, return 0.

Clarifying Questions

  1. What do we mean by “picking optimally”?
    • Both players aim to maximize their own coin values at each turn.
  2. Should the players always alternate turns starting from the first player?
    • Yes, the players alternate turns starting with the first player.
  3. What is the length of the input array?
    • There is no restriction provided on the length of the array; we need to assume it can be of any length.

Strategy

This is a typical game theory problem that can be efficiently solved using dynamic programming. The goal is to calculate two values:

We will use a bottom-up dynamic programming approach to solve this:

  1. Define dp[left][right] as the maximum value the current player can collect from coins[left:right+1].
  2. Use a 2D DP table where left and right define a subarray from left to right in coins.
  3. Base case: When left == right, the player can only pick one coin, i.e., coins[left].

Recurrence relation:

Code

Let’s implement this in Python:

def find_winning_player(coins):
    n = len(coins)
    if n == 0:
        return 0

    # Initializing dp array and sum array
    dp = [[0]*n for _ in range(n)]
    sum_coins = [0]*(n+1)

    # Compute the prefix sum
    for i in range(n):
        sum_coins[i+1] = sum_coins[i] + coins[i]

    # Base cases
    for i in range(n):
        dp[i][i] = coins[i]

    # Dynamic programming to fill dp table
    for length in range(2, n+1):
        for left in range(n - length + 1):
            right = left + length - 1
            dp[left][right] = max(
                coins[left] + (sum_coins[right+1] - sum_coins[left+1]) - dp[left+1][right],
                coins[right] + (sum_coins[right] - sum_coins[left]) - dp[left][right-1]
            )
    
    # Final computation of sums
    total_sum = sum(coins)
    first_player = dp[0][n-1]
    second_player = total_sum - first_player
    
    if first_player > second_player:
        return 1
    elif second_player > first_player:
        return 2
    else:
        return 0

# Example
coins = [1, 5, 2]
print(find_winning_player(coins))  # Output should be 1 or 2 depending on who can optimize better.

Time Complexity

The time complexity of this approach is O(n^2), where n is the length of the coins array. This is because we are building up solutions for all subarrays of coins, and each subarray computation involves a constant amount of work.

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