Solution Approach
Solution Overview
There are 2 main approaches to solve this problem:
- Brute Force Approach - Complex approach
- Greedy Approach with Sorting - Complex approach
Approach 1: Brute Force Approach
Let's start by understanding the problem: we have tokens with different values, and we need to maximize our score by playing them strategically.
Thinking Process: The most straightforward approach would be to try all possible combinations of playing tokens face up or face down, and find the one that gives the maximum score.
For each token, we have three choices: play it face up, play it face down, or don't play it at all. This leads to 3^n possible combinations, where n is the number of tokens.
Intuition: This approach directly follows the problem statement by exploring all possible ways to play the tokens. However, it's extremely inefficient due to the exponential number of combinations.
Algorithm:
- Generate all possible combinations of playing tokens (face up, face down, or not playing).
- For each combination, simulate the game and calculate the final score.
- Return the maximum score achieved across all combinations.
Time Complexity:
Where n is the number of tokens. We have three choices for each token, leading to 3^n possible combinations.
Space Complexity:
We need O(n) space to store the current combination of choices.
Approach 2: Greedy Approach with Sorting
We can optimize the brute force approach by recognizing that there's a greedy strategy that always leads to the optimal solution.
Thinking Process: The key insight is that if we're going to play tokens face up, we should play the ones with the smallest values first to minimize power loss. And if we're going to play tokens face down, we should play the ones with the largest values first to maximize power gain.
So, we can sort the tokens by their values. Then, we can use two pointers: one pointing to the smallest token (left) and one pointing to the largest token (right). At each step, we have three choices:
- Play the smallest token face up if we have enough power.
- Play the largest token face down if we have at least 1 score.
- Stop playing if neither of the above is possible.
Intuition: This greedy approach works because the optimal strategy is to always play the smallest tokens face up and the largest tokens face down. By sorting the tokens and using two pointers, we can efficiently implement this strategy.
Algorithm:
- Sort the tokens in ascending order.
- Initialize two pointers: left pointing to the smallest token and right pointing to the largest token.
- Initialize variables to track the current score and the maximum score achieved.
- While the left pointer is less than or equal to the right pointer:
- a. If we have enough power to play the smallest token face up, do so and increment the left pointer and the current score.
- b. Else if we have at least 1 score, play the largest token face down, decrement the right pointer and the current score.
- c. Else, break the loop as we can't make any more moves.
- Return the maximum score achieved.
Time Complexity:
Where n is the number of tokens. Sorting the tokens takes O(n log n) time, and the two-pointer approach takes O(n) time.
Space Complexity:
We only need a constant amount of extra space for the pointers and score variables.
Pseudocode
12345678910111213141516171819202122function bagOfTokensScore(tokens, power): n = tokens.length maxScore = 0 function backtrack(index, currentPower, currentScore): if index == n: maxScore = max(maxScore, currentScore) return // Don't play this token backtrack(index + 1, currentPower, currentScore) // Play face up if possible if currentPower >= tokens[index]: backtrack(index + 1, currentPower - tokens[index], currentScore + 1) // Play face down if possible if currentScore >= 1: backtrack(index + 1, currentPower + tokens[index], currentScore - 1) backtrack(0, power, 0) return maxScoreString Problems
Learning Objective
Understand different approaches to solve the token strategy game problem and analyze their efficiency.
Solution Approaches
Approach 1: Brute Force Approach
Let's start by understanding the problem: we have tokens with different values, and we need to maximize our score by playing them strategically.
Thinking Process: The most straightforward approach would be to try all possible combinations of playing tokens face up or face down, and find the one that gives the maximum score.
For each token, we have three choices: play it face up, play it face down, or don't play it at all. This leads to 3^n possible combinations, where n is the number of tokens.
Intuition: This approach directly follows the problem statement by exploring all possible ways to play the tokens. However, it's extremely inefficient due to the exponential number of combinations.
Algorithm:
- Generate all possible combinations of playing tokens (face up, face down, or not playing).
- For each combination, simulate the game and calculate the final score.
- Return the maximum score achieved across all combinations.
Approach 2: Greedy Approach with Sorting
We can optimize the brute force approach by recognizing that there's a greedy strategy that always leads to the optimal solution.
Thinking Process: The key insight is that if we're going to play tokens face up, we should play the ones with the smallest values first to minimize power loss. And if we're going to play tokens face down, we should play the ones with the largest values first to maximize power gain.
So, we can sort the tokens by their values. Then, we can use two pointers: one pointing to the smallest token (left) and one pointing to the largest token (right). At each step, we have three choices:
- Play the smallest token face up if we have enough power.
- Play the largest token face down if we have at least 1 score.
- Stop playing if neither of the above is possible.
Intuition: This greedy approach works because the optimal strategy is to always play the smallest tokens face up and the largest tokens face down. By sorting the tokens and using two pointers, we can efficiently implement this strategy.
Algorithm:
- Sort the tokens in ascending order.
- Initialize two pointers: left pointing to the smallest token and right pointing to the largest token.
- Initialize variables to track the current score and the maximum score achieved.
- While the left pointer is less than or equal to the right pointer:
- a. If we have enough power to play the smallest token face up, do so and increment the left pointer and the current score.
- b. Else if we have at least 1 score, play the largest token face down, decrement the right pointer and the current score.
- c. Else, break the loop as we can't make any more moves.
- Return the maximum score achieved.
Complexity Analysis
Brute Force Approach Approach
Time Complexity
Where n is the number of tokens. We have three choices for each token, leading to 3^n possible combinations.
Space Complexity
We need O(n) space to store the current combination of choices.
Greedy Approach with Sorting Approach
Time Complexity
Where n is the number of tokens. Sorting the tokens takes O(n log n) time, and the two-pointer approach takes O(n) time.
Space Complexity
We only need a constant amount of extra space for the pointers and score variables.
Pseudocode
Brute Force Approach Approach
12345678910111213141516171819202122function bagOfTokensScore(tokens, power): n = tokens.length maxScore = 0 function backtrack(index, currentPower, currentScore): if index == n: maxScore = max(maxScore, currentScore) return // Don't play this token backtrack(index + 1, currentPower, currentScore) // Play face up if possible if currentPower >= tokens[index]: backtrack(index + 1, currentPower - tokens[index], currentScore + 1) // Play face down if possible if currentScore >= 1: backtrack(index + 1, currentPower + tokens[index], currentScore - 1) backtrack(0, power, 0) return maxScoreGreedy Approach with Sorting Approach
1234567891011121314151617181920212223242526function bagOfTokensScore(tokens, power): // Sort tokens in ascending order sort(tokens) left = 0 right = tokens.length - 1 score = 0 maxScore = 0 while left <= right: // If we can play the smallest token face up if power >= tokens[left]: power -= tokens[left] left++ score++ maxScore = max(maxScore, score) // If we can play the largest token face down else if score > 0: power += tokens[right] right-- score-- // If we can't make any move else: break return maxScore