Problem Description

You have n players numbered from 0 to n-1 playing a ball-passing game. Each player i has a predetermined receiver stored in receiver[i], which indicates the player they will pass the ball to.

The game works as follows:

• You choose any starting player i
• Player i passes the ball to player receiver[i]
• Player receiver[i] then passes to player receiver[receiver[i]]
• This continues for exactly k passes total

The score of a game is calculated as the sum of all player indices who touched the ball during these k passes, including the starting player. If a player touches the ball multiple times, their index is counted each time.

For example, if the ball goes through players i → receiver[i] → receiver[receiver[i]] → ... for k passes, the score would be i + receiver[i] + receiver[receiver[i]] + ... (summing k+1 terms total, since we include the starting player).

Important notes:

• The receiver array may contain duplicates (multiple players can pass to the same player)
• A player can pass to themselves (receiver[i] can equal i)

Your task is to find the maximum possible score by choosing the optimal starting player.

Intuition

The naive approach would be to simulate the ball passing for each starting player, following the path for k passes and calculating the sum. However, with k potentially being very large (up to 10^10 based on typical constraints), this would result in O(n * k) time complexity, which is too slow.

The key observation is that we're essentially following a path in a directed graph where each node has exactly one outgoing edge. Starting from any player i, we need to find the sum of indices along a path of length k.

When dealing with problems that require traversing a large number of steps along a fixed path, binary lifting is a powerful technique. The idea is to precompute jumps of powers of 2: instead of moving one step at a time, we can jump 2^0, 2^1, 2^2, ... steps.

Why does this help? Any number k can be represented as a sum of powers of 2 (its binary representation). For example, if k = 13 = 1101₂ = 8 + 4 + 1, we can decompose the k passes into jumps of 8 passes, 4 passes, and 1 pass.

For each player i and jump size 2^j, we need to track two things:

1. Where do we end up after 2^j passes starting from player i - this helps us chain jumps together
2. What's the sum of indices for those 2^j passes (excluding the final position to avoid double counting when chaining)

By building these tables using dynamic programming (where 2^j passes = 2^(j-1) passes followed by another 2^(j-1) passes), we can then reconstruct any path of length k by combining the appropriate power-of-2 jumps based on k's binary representation. This reduces our time complexity from O(k) per starting player to O(log k) per starting player.

Learn more about Dynamic Programming patterns.

Solution Approach

We implement the solution using dynamic programming with binary lifting. Let's break down the implementation step by step:

1. Initialize the DP Tables

We create two 2D arrays:

• f[i][j]: The player we reach after making 2^j passes starting from player i
• g[i][j]: The sum of player indices visited during 2^j passes starting from player i (excluding the final player to avoid double counting)

The dimensions are n × m where n is the number of players and m = k.bit_length() (the number of bits needed to represent k).

2. Base Case (j = 0)

For single passes (2^0 = 1 pass):

• f[i][0] = receiver[i]: After 1 pass from player i, we reach receiver[i]
• g[i][0] = i: The sum includes only the starting player i (not the receiver)

3. Build the DP Tables

For j > 0, we compute 2^j passes by combining two 2^(j-1) passes:

• f[i][j] = f[f[i][j-1]][j-1]: First jump 2^(j-1) steps to reach f[i][j-1], then jump another 2^(j-1) steps from there
• g[i][j] = g[i][j-1] + g[f[i][j-1]][j-1]: The sum is the sum from the first half plus the sum from the second half

4. Calculate Maximum Score

For each potential starting player i:

• Initialize position p = i and sum t = 0
• Decompose k into its binary representation
• For each bit j that is set in k (i.e., k >> j & 1 is true):
  • Add g[p][j] to the running sum t
  • Update position to p = f[p][j]
• After processing all bits, add the final position p to get the total score
• Track the maximum score across all starting players

Example:

If k = 5 = 101₂, we need to make jumps of 2^0 = 1 and 2^2 = 4 passes. Starting from player i:

• Jump 4 passes: accumulate g[i][2], move to f[i][2]
• Jump 1 pass: accumulate g[current_position][0], move to f[current_position][0]
• Add the final position to get the complete sum

The time complexity is O(n × log k) for building the tables and finding the maximum, and space complexity is O(n × log k) for storing the DP tables.

Example Walkthrough

Let's walk through a concrete example to illustrate the binary lifting solution.

Given:

• n = 4 players (indexed 0, 1, 2, 3)
• receiver = [1, 2, 3, 0] (player 0 passes to 1, player 1 passes to 2, etc.)
• k = 5 passes

Step 1: Build the DP tables

First, let's understand what paths look like from each player:

• From player 0: 0 → 1 → 2 → 3 → 0 → 1 → ...
• From player 1: 1 → 2 → 3 → 0 → 1 → 2 → ...
• From player 2: 2 → 3 → 0 → 1 → 2 → 3 → ...
• From player 3: 3 → 0 → 1 → 2 → 3 → 0 → ...

Since k = 5 = 101₂, we need tables for j = 0, 1, 2 (powers of 2 up to 4).

Base case (j = 0, representing 2^0 = 1 pass):

• f[0][0] = 1, g[0][0] = 0 (from 0, one pass reaches 1, sum = 0)
• f[1][0] = 2, g[1][0] = 1 (from 1, one pass reaches 2, sum = 1)
• f[2][0] = 3, g[2][0] = 2 (from 2, one pass reaches 3, sum = 2)
• f[3][0] = 0, g[3][0] = 3 (from 3, one pass reaches 0, sum = 3)

For j = 1 (representing 2^1 = 2 passes):

• f[0][1] = f[f[0][0]][0] = f[1][0] = 2 (0 → 1 → 2)
• g[0][1] = g[0][0] + g[1][0] = 0 + 1 = 1 (sum of indices 0, 1)
• f[1][1] = f[f[1][0]][0] = f[2][0] = 3 (1 → 2 → 3)
• g[1][1] = g[1][0] + g[2][0] = 1 + 2 = 3 (sum of indices 1, 2)
• Similarly: f[2][1] = 0, g[2][1] = 5 and f[3][1] = 1, g[3][1] = 3

For j = 2 (representing 2^2 = 4 passes):

• f[0][2] = f[f[0][1]][1] = f[2][1] = 0 (0 → 1 → 2 → 3 → 0)
• g[0][2] = g[0][1] + g[2][1] = 1 + 5 = 6 (sum of indices 0, 1, 2, 3)
• Similarly for other players...

Step 2: Calculate score for each starting player

Let's trace through starting from player 0 with k = 5 = 101₂:

1. Initialize: position = 0, sum = 0
2. Process bit 0 (2^0 = 1 pass):
   • Add g[0][0] = 0 to sum → sum = 0
   • Move to f[0][0] = 1 → position = 1
3. Skip bit 1 (not set in 101₂)
4. Process bit 2 (2^2 = 4 passes):
   • Add g[1][2] = 7 to sum → sum = 7
   • Move to f[1][2] = 1 → position = 1
5. Add final position: sum = 7 + 1 = 8

The path taken: 0 → 1 → 2 → 3 → 0 → 1 Score = 0 + 1 + 2 + 3 + 0 + 1 = 7... Wait, let me recalculate.

Actually, with 5 passes starting from player 0:

• Pass 1: 0 → 1 (ball at 1)
• Pass 2: 1 → 2 (ball at 2)
• Pass 3: 2 → 3 (ball at 3)
• Pass 4: 3 → 0 (ball at 0)
• Pass 5: 0 → 1 (ball at 1)

Players who touched the ball: 0, 1, 2, 3, 0, 1 Score = 0 + 1 + 2 + 3 + 0 + 1 = 7

Step 3: Compare all starting positions

• Starting from 0: Score = 7
• Starting from 1: Score = 8 (path: 1 → 2 → 3 → 0 → 1 → 2, sum = 1+2+3+0+1+2 = 9)
• Starting from 2: Score = 10 (path: 2 → 3 → 0 → 1 → 2 → 3, sum = 2+3+0+1+2+3 = 11)
• Starting from 3: Score = 9 (path: 3 → 0 → 1 → 2 → 3 → 0, sum = 3+0+1+2+3+0 = 9)

The maximum score is 11, achieved by starting from player 2.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * log k)

The algorithm uses binary lifting technique where:

• Building the jump tables requires two nested loops: the outer loop runs m = k.bit_length() times (which is O(log k)), and the inner loop runs n times for each iteration. This gives us O(n * log k) for preprocessing.
• Computing the answer requires iterating through all n starting positions, and for each position, we iterate through at most m = O(log k) bits of k to perform the jumps. This also gives us O(n * log k).
• Overall time complexity is O(n * log k) + O(n * log k) = O(n * log k).

Space Complexity: O(n * log k)

The algorithm maintains two 2D arrays:

• Array f of size n × m where m = k.bit_length() = O(log k), storing the destination after 2^j jumps from each position.
• Array g of size n × m, storing the sum of values collected after 2^j jumps from each position.
• Both arrays require O(n * log k) space.
• Additional variables use O(1) space.
• Overall space complexity is O(n * log k).

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

Common Pitfalls

1. Off-by-One Error in Counting Nodes

The Pitfall: A critical misunderstanding is how many nodes are involved when making k passes. If we make k passes, we actually touch k+1 nodes total (the starting node plus k nodes we pass through). Many implementations incorrectly assume only k nodes are involved.

Why This Happens: The problem states "k passes" but the score includes the starting player. When player i makes a pass to receiver[i], that's 1 pass but 2 players touched the ball.

The Fix: The provided solution handles this correctly by:

• Storing partial sums in sum_values[i][j] that exclude the final destination
• Adding the final node (current_node) separately after all jumps: function_value = total_sum + current_node

2. Integer Overflow for Large k Values

The Pitfall: When k is very large (up to 10^10 as per typical constraints), and we're summing player indices multiple times, the total sum can exceed standard integer limits.

Example: If k = 10^10 and players keep passing in a small cycle like 0 → 1 → 0, the sum could be approximately 5 × 10^9, which exceeds 32-bit integer limits.

The Fix:

• In Python, integers automatically handle arbitrary precision, so this isn't an issue
• In languages like Java/C++, use 64-bit integers (long in Java, long long in C++)

3. Incorrect Binary Decomposition Logic

The Pitfall: When decomposing k into powers of 2, developers might accidentally process bits in the wrong order or forget to update the current position correctly.

Wrong approach example:

# INCORRECT: Forgetting to update position
for bit in range(max_power):
    if (k >> bit) & 1:
        total_sum += sum_values[start_node][bit]  # Always using start_node!

The Fix: Always update the current position after each jump:

for bit_position in range(max_power):
    if (k >> bit_position) & 1:
        total_sum += sum_values[current_node][bit_position]
        current_node = next_node[current_node][bit_position]  # Update position!

4. Misunderstanding the Sum Storage in Binary Lifting

The Pitfall: Storing the wrong values in sum_values[i][j]. Some might think it should include the destination node, leading to double-counting.

Wrong interpretation: sum_values[i][j] = sum of all 2^j + 1 nodes (including both start and end)

Correct interpretation: sum_values[i][j] = sum of the first 2^j nodes (excluding the final destination)

Why This Matters: When combining two jumps of size 2^(j-1), the intermediate node would be counted twice if we included destinations in our sums.

5. Edge Case: k = 0

The Pitfall: The problem might allow k = 0 (no passes), which means only the starting player touches the ball.

The Fix: The current solution handles this correctly because:

• When k = 0, no bits are set, so the loop doesn't execute
• function_value = total_sum + current_node = 0 + start_node, which is correct

However, it's worth adding an explicit check for clarity:

if k == 0:
    return max(range(n))  # Maximum player index