#2836
Maximize Value of Function in a Ball Passing Game
HardArrayDynamic ProgrammingBit ManipulationDynamic ProgrammingBinary LiftingGraph Traversal
Approaches
Brute ForceOptimal
Complexity Comparison
| Brute Force | Optimal Solution★ | |
|---|---|---|
| Time | O(n²) | O(n log k) |
| Space | O(1) | O(n log k) |
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Intuition
Time O(n log k)Space O(n log k)
The optimal solution uses a technique called binary lifting to efficiently compute the score after k passes. Instead of simulating each pass, we can jump directly to the player who would receive the ball after a certain number of passes, reducing the time complexity significantly.
⚙️
Algorithm
3 steps- 1Step 1: Precompute the next player for each player using binary lifting.
- 2Step 2: For each player, calculate the total score for k passes using the precomputed values.
- 3Step 3: Return the maximum score found.
solution.py20 lines
1def maxScore(receiver, k):
2 n = len(receiver)
3 max_score = 0
4 log_k = k.bit_length()
5 jump = [[0] * log_k for _ in range(n)]
6 for i in range(n):
7 jump[i][0] = receiver[i]
8 for j in range(1, log_k):
9 for i in range(n):
10 jump[i][j] = jump[jump[i][j-1]][j-1]
11 for i in range(n):
12 current_score = 0
13 current_player = i
14 for j in range(log_k):
15 if k & (1 << j):
16 current_score += current_player
17 current_player = jump[current_player][j]
18 current_score += current_player
19 max_score = max(max_score, current_score)
20 return max_scoreℹ
Complexity note: The time complexity is O(n log k) because we preprocess the jump table in O(n log k) and then compute scores in O(n log k) as well, making it efficient even for large k.
- 1Understanding the relationship between players and how the ball is passed is crucial for optimizing the solution.
- 2Binary lifting allows us to efficiently compute the result without simulating every pass.
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