LeetCode #2836 — HARD

Maximize Value of Function in a Ball Passing Game

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array receiver of length n and an integer k. n players are playing a ball-passing game.

You choose the starting player, i. The game proceeds as follows: player i passes the ball to player receiver[i], who then passes it to receiver[receiver[i]], and so on, for k passes in total. The game's score is the sum of the indices of the players who touched the ball, including repetitions, i.e. i + receiver[i] + receiver[receiver[i]] + ... + receiver^(k)[i].

Return the maximum possible score.

Notes:

receiver may contain duplicates.
receiver[i] may be equal to i.

Example 1:

Input: receiver = [2,0,1], k = 4

Output: 6

Explanation:

Starting with player i = 2 the initial score is 2:

Pass	Sender Index	Receiver Index	Score
1	2	1	3
2	1	0	3
3	0	2	5
4	2	1	6

Example 2:

Input: receiver = [1,1,1,2,3], k = 3

Output: 10

Explanation:

Starting with player i = 4 the initial score is 4:

Pass	Sender Index	Receiver Index	Score
1	4	3	7
2	3	2	9
3	2	1	10

Constraints:

1 <= receiver.length == n <= 10⁵
0 <= receiver[i] <= n - 1
1 <= k <= 10¹⁰

Step 01

Brute Force Baseline

Problem summary: You are given an integer array receiver of length n and an integer k. n players are playing a ball-passing game. You choose the starting player, i. The game proceeds as follows: player i passes the ball to player receiver[i], who then passes it to receiver[receiver[i]], and so on, for k passes in total. The game's score is the sum of the indices of the players who touched the ball, including repetitions, i.e. i + receiver[i] + receiver[receiver[i]] + ... + receiver(k)[i]. Return the maximum possible score. Notes: receiver may contain duplicates. receiver[i] may be equal to i.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Bit Manipulation

Example 1

[2,0,1]
4

Example 2

[1,1,1,2,3]
3

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">We can solve the problem using binary lifting.</div>
<div class="_1l1MA">For each player with id <code>x</code> and for every <code>i</code> in the range <code>[0, ceil(log<sub>2</sub>k)]</code>, we can determine the last receiver's id and compute the sum of player ids who receive the ball after <code>2<sup>i</sup></code> passes, starting from <code>x</code>.</div>
<div class="_1l1MA">Let <code>last_receiver[x][i] =</code> the last receiver's id after <code>2<sup>i</sup></code> passes, and <code>sum[x][i] =</code> the sum of player ids who receive the ball after <code>2<sup>i</sup></code> passes. For all <code>x</code> in the range <code>[0, n - 1]</code>, <code>last_receiver[x][0] = receiver[x]</code>, and <code>sum[x][0] = receiver[x]</code>.</div>
<div class="_1l1MA">Then for <code>i</code> in range <code>[1, ceil(log<sub>2</sub>k)]</code>, <code>last_receiver[x][i] = last_receiver[last_receiver[x][i - 1]][i - 1]</code> and <code>sum[x][i] = sum[x][i - 1] + sum[last_receiver[x][i - 1]][i - 1]</code>, for all <code>x</code> in the range <code>[0, n - 1]</code>.</div>
<div class="_1l1MA">Starting from each player id <code>x</code>, we can now go through the powers of <code>2</code> in the binary representation of <code>k</code> and make jumps corresponding to each power, using the pre-computed values, to compute <code>f(x)</code>.</div>
<div class="_1l1MA">The answer is the maximum <code>f(x)</code> from each player id.</div>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
class Solution {
    public long getMaxFunctionValue(List<Integer> receiver, long k) {
        int n = receiver.size(), m = 64 - Long.numberOfLeadingZeros(k);
        int[][] f = new int[n][m];
        long[][] g = new long[n][m];
        for (int i = 0; i < n; ++i) {
            f[i][0] = receiver.get(i);
            g[i][0] = i;
        }
        for (int j = 1; j < m; ++j) {
            for (int i = 0; i < n; ++i) {
                f[i][j] = f[f[i][j - 1]][j - 1];
                g[i][j] = g[i][j - 1] + g[f[i][j - 1]][j - 1];
            }
        }
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            int p = i;
            long t = 0;
            for (int j = 0; j < m; ++j) {
                if ((k >> j & 1) == 1) {
                    t += g[p][j];
                    p = f[p][j];
                }
            }
            ans = Math.max(ans, p + t);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
func getMaxFunctionValue(receiver []int, k int64) (ans int64) {
	n, m := len(receiver), bits.Len(uint(k))
	f := make([][]int, n)
	g := make([][]int64, n)
	for i := range f {
		f[i] = make([]int, m)
		g[i] = make([]int64, m)
		f[i][0] = receiver[i]
		g[i][0] = int64(i)
	}
	for j := 1; j < m; j++ {
		for i := 0; i < n; i++ {
			f[i][j] = f[f[i][j-1]][j-1]
			g[i][j] = g[i][j-1] + g[f[i][j-1]][j-1]
		}
	}
	for i := 0; i < n; i++ {
		p := i
		t := int64(0)
		for j := 0; j < m; j++ {
			if k>>j&1 == 1 {
				t += g[p][j]
				p = f[p][j]
			}
		}
		ans = max(ans, t+int64(p))
	}
	return
}

# Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
class Solution:
    def getMaxFunctionValue(self, receiver: List[int], k: int) -> int:
        n, m = len(receiver), k.bit_length()
        f = [[0] * m for _ in range(n)]
        g = [[0] * m for _ in range(n)]
        for i, x in enumerate(receiver):
            f[i][0] = x
            g[i][0] = i
        for j in range(1, m):
            for i in range(n):
                f[i][j] = f[f[i][j - 1]][j - 1]
                g[i][j] = g[i][j - 1] + g[f[i][j - 1]][j - 1]
        ans = 0
        for i in range(n):
            p, t = i, 0
            for j in range(m):
                if k >> j & 1:
                    t += g[p][j]
                    p = f[p][j]
            ans = max(ans, t + p)
        return ans

// Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
// class Solution {
//     public long getMaxFunctionValue(List<Integer> receiver, long k) {
//         int n = receiver.size(), m = 64 - Long.numberOfLeadingZeros(k);
//         int[][] f = new int[n][m];
//         long[][] g = new long[n][m];
//         for (int i = 0; i < n; ++i) {
//             f[i][0] = receiver.get(i);
//             g[i][0] = i;
//         }
//         for (int j = 1; j < m; ++j) {
//             for (int i = 0; i < n; ++i) {
//                 f[i][j] = f[f[i][j - 1]][j - 1];
//                 g[i][j] = g[i][j - 1] + g[f[i][j - 1]][j - 1];
//             }
//         }
//         long ans = 0;
//         for (int i = 0; i < n; ++i) {
//             int p = i;
//             long t = 0;
//             for (int j = 0; j < m; ++j) {
//                 if ((k >> j & 1) == 1) {
//                     t += g[p][j];
//                     p = f[p][j];
//                 }
//             }
//             ans = Math.max(ans, p + t);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2836: Maximize Value of Function in a Ball Passing Game
// class Solution {
//     public long getMaxFunctionValue(List<Integer> receiver, long k) {
//         int n = receiver.size(), m = 64 - Long.numberOfLeadingZeros(k);
//         int[][] f = new int[n][m];
//         long[][] g = new long[n][m];
//         for (int i = 0; i < n; ++i) {
//             f[i][0] = receiver.get(i);
//             g[i][0] = i;
//         }
//         for (int j = 1; j < m; ++j) {
//             for (int i = 0; i < n; ++i) {
//                 f[i][j] = f[f[i][j - 1]][j - 1];
//                 g[i][j] = g[i][j - 1] + g[f[i][j - 1]][j - 1];
//             }
//         }
//         long ans = 0;
//         for (int i = 0; i < n; ++i) {
//             int p = i;
//             long t = 0;
//             for (int j = 0; j < m; ++j) {
//                 if ((k >> j & 1) == 1) {
//                     t += g[p][j];
//                     p = f[p][j];
//                 }
//             }
//             ans = Math.max(ans, p + t);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log k)

Space

O(n × log k)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.