2836. Maximize Value of Function in a Ball Passing Game

Problem Description

In this problem, you are provided with an array receiver of length n representing a list of n players (each having a unique identifier from 0 to n - 1) participating in a game where they pass a ball to each other. The ball passing is described by receiver[i], which indicates the player that receives the ball from player i. Importantly, a player can pass the ball to themselves.

You are also given an integer k, which is the exact number of times the ball will be passed starting from any chosen player. Your goal is to select a starting player such that the sum of the ids of all players who touch the ball (f(x)), including repetitions and the starting player's id, is maximized after exactly k passes.

To clarify, f(x) is the function that calculates the total mentioned above where x is the starting player's id, and its formula is f(x) = x + receiver[x] + receiver[receiver[x]] + ... until k terms are added up.

Your task is to determine the maximum value that f(x) can reach for any starting player's id x and return this maximum value.

Intuition

The solution involves using dynamic programming to efficiently calculate the sum of ids for each potential starting player, given k passes.

First, we observe that calculating f(x) directly could be costly if k is large, as it would involve following the chain of passes k times. Instead, we can precompute the pass receivers and the id sums for different powers of two, which can be represented in the binary form of k. This allows us to decompose k into a sum of powers of two and compute f(x) using these precomputed values, hence reducing the time complexity.

The idea is to create two matrices f and g, where f[i][j] stores the id of the player who receives the ball from player i after 2^j passes and g[i][j] stores the sum of player ids in 2^j passes from player i. These matrices are filled using dynamic programming.

Once these matrices are precomputed, we iterate over each player i as the starting player. We decompose k into its binary representation and simulate the ball passing to accumulate the sum of ids by checking each bit of k. If the j-th bit of k is set, we update our current player to f[current_player][j] and add the corresponding ids from g[current_player][j] to our total sum.

Utilizing the precomputed values from matrices f and g, we find the accumulated sum of ids t for each possible starting player, and we keep track of the maximum value encountered. The final answer is the largest accumulated sum t plus the last player's id that would be reached with the k passes from each starting player.

The given solution code implements this approach to efficiently compute the desired maximum function value f(x) for a given receiver array and a number of passes k.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation of the solution leverages dynamic programming and bit manipulation concepts. Here is a step-by-step breakdown of the approach used in the provided Python code:

1. Initialization: Two matrix data structures f and g of dimensions n x m are initialized. Here, n is the number of players and m is the bit-length of k (which represents the number of powers of 2 required to express k in binary form). f[i][j] will ultimately represent which player has the ball after 2^j passes starting from player i, while g[i][j] will store the sum of player ids for 2^j passes from player i.

2. Base Case for DP:

   • The zeroth column of f, which corresponds to 2^0 (or 1 pass), is filled with receiver[i] indicating the direct receiver of the ball from each player.
   • The zeroth column of g is filled with i since the sum of ids after 1 pass from the starting player i is just i.

3. DP Recurrence:

   • The matrices are filled using a recurrence relation. For a given j-iteration (j going from 1 to m-1), f[i][j] is populated with the player who will receive from the player at f[i][j-1] after 2^(j-1) passes; thus it's like making a "jump" of 2^(j-1) twice, leading to a total of 2^j passes.
   • Similarly, g[i][j] is the sum of ids for 2^j passes and is calculated as the sum of ids for 2^(j-1) passes from player i and from player f[i][j-1].

4. Calculating the Maximum Value of f(x):

   • The code iterates over each player i considering them as the starting player. For each player, it initializes p as the current player and t as the total sum of ids.
   • It then goes through the bits of k. If the j-th bit of k is set, it means 2^j passes are part of the total k passes. The current player is updated, and the corresponding sum of ids is added to the total.
   • After considering all set bits in k, the final id p is added to the total t. This gives the value of f(i) where i is the starting player.

5. Result:

   • The maximum value obtained for t + p across all starting players i is the result of the function and is returned.

The dynamic programming aspect of this solution significantly reduces the amount of computation needed by dividing the problem into subproblems that can reuse solutions. It optimizes the process of figuring out the result of k passes by using the binary representation of k and avoids explicit linear chaining of passes.

The recursive relationships in the f and g matrices and bit manipulation for calculating the pass sums are the core components used to derive the solution.

Example Walkthrough

Let's illustrate the solution approach with a small example.

Suppose we have the following receiver array representing the players who get the ball next:

1receiver = [1, 2, 3, 0]

This means that player 0 passes the ball to player 1, player 1 passes it to player 2, and so on. If a player would pass the ball back to themselves, it would be indicated by having their own index in their slot (not the case here).

Let's also assume k = 3, meaning the ball will be passed exactly 3 times. Our task is to maximize the sum f(x) of the player ids over these 3 passes.

Step 1: Initialize the matrices f and g. Our matrix will be 4 x 2 since k = 3 (011 in binary, so we need two bits to express it).

Step 2: Fill the base case in both matrices.

• For f[i][0], we have f[0][0] = 1, f[1][0] = 2, f[2][0] = 3, and f[3][0] = 0.
• For g[i][0], it is straightforward: g[i][0] = i, so g[0][0] = 0, g[1][0] = 1, etc.

Step 3: Fill the matrices using the dynamic programming approach.

• We observe that when receiver[i] == receiver[receiver[i]] (which doesn’t happen in this case), we can set f[i][1] directly to i, otherwise use the formula: f[i][1] = receiver[receiver[i]]. So, f[0][1] = f[1][0] = 2, f[1][1] = f[2][0] = 3, etc.
• For g, we have g[i][1] = g[i][0] + g[f[i][0]][0], leading to g[0][1] = 0 + 1 = 1, g[1][1] = 1 + 2 = 3, etc.

Step 4: Calculate the maximum value of f(x) by iterating over each player as the starting player and considering the bits of k.

• Initially, t = 0 and p is the starting player.
• For each player i as the starting player, examine the bits of k, updating p and adding to t accordingly. For k = 3, we have two binary digits to consider (11).
• Let's choose player 0 as the starting player. The first bit (rightmost) is 1, so we add g[0][0] to t and update p to f[0][0]. Now t = 0 and p = 1.
• The second bit (next left) is also 1, so we add g[1][1] to t and update p to f[1][1]. Now t = 3 and p = 3.
• Finally, add player 3's id to t: t = 3 + 3 = 6.
• We would iterate this process for each player to find the maximum sum.

Step 5: Result

• The maximum value of t + p that we computed for each starting player is found. In this case, starting with player 0 gave us 6. If we repeat this for all players, we might find that 6 is the maximum or find a higher sum. Let’s say it remains the maximum sum after we went through each player; hence, 6 is our result.

As seen in this small example, the method utilizes dynamic programming to simplify the computation and avoid direct k-pass simulation by instead utilizing a binary representation and matrix table lookups. This makes the solution more efficient, especially for large values of k.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity comprises three parts: initializing the 2D arrays f and g, filling in the sparse table f and g, and calculating the maximum function value ans.

1. Initializing the 2D arrays:

   • Initializing two 2D arrays, f and g, of size n x m each, where n is the length of the receiver list and m is the bit_length() of k. This initialization is done with two list comprehensions that have a complexity of O(n * m).

2. Filling the sparse table:

   • A nested loop is used where the outer loop runs for m iterations (the number of bits in k) and the inner loop runs for n iterations (the length of the receiver). Thus, the complexity for this part is O(n * m).

3. Calculating the maximum function value:

   • A nested loop where the outer loop runs for n iterations and the inner loop, conditionally executed based on the bits of k, runs up to m times. So, in the worst case, this part has a complexity of O(n * m).

Summing them all up, the total time complexity is O(n * m + n * m + n * m) = O(n * m).

Space Complexity

The space complexity is determined by the space taken up by the 2D arrays f and g, and by any additional variables used.

1. Space for 2D arrays:

   • Two 2D arrays f and g, each of size n x m, give a space complexity of O(n * m).

2. Additional variables:

   • Variables n, m, i, j, p, t, ans, and a few others are used. However, these do not significantly affect the space complexity as their space requirement is constant.

Consequently, the total space complexity is O(n * m).

Learn more about how to find time and space complexity quickly using problem constraints.