Bag of Tokens

Description

You start with an initial power of power, an initial score of 0, and a bag of tokens given as an integer array tokens, where each tokens[i] donates the value of token_i_.

Your goal is to maximize the total score by strategically playing these tokens. In one move, you can play an unplayed token in one of the two ways (but not both for the same token):

• Face-up: If your current power is at least tokens[i], you may play token_i_, losing tokens[i] power and gaining 1 score.
• Face-down: If your current score is at least 1, you may play token_i_, gaining tokens[i] power and losing 1 score.

Return the maximum possible score you can achieve after playing any number of tokens.

Example 1:

Input: tokens = [100], power = 50

Output: 0

Explanation**:** Since your score is 0 initially, you cannot play the token face-down. You also cannot play it face-up since your power (50) is less than tokens[0] (100).

Example 2:

Input: tokens = [200,100], power = 150

Output: 1

Explanation: Play token_1_ (100) face-up, reducing your power to 50 and increasing your score to 1.

There is no need to play token_0_, since you cannot play it face-up to add to your score. The maximum score achievable is 1.

Example 3:

Input: tokens = [100,200,300,400], power = 200

Output: 2

Explanation: Play the tokens in this order to get a score of 2:

1. Play token_0_ (100) face-up, reducing power to 100 and increasing score to 1.
2. Play token_3_ (400) face-down, increasing power to 500 and reducing score to 0.
3. Play token_1_ (200) face-up, reducing power to 300 and increasing score to 1.
4. Play token_2_ (300) face-up, reducing power to 0 and increasing score to 2.

The maximum score achievable is 2.

Constraints:

• 0 <= tokens.length <= 1000
• 0 <= tokens[i], power < 104

Code

Time Complexity: $O(n\log n)$, Space Complexity: $O(1)$

Greedy Intuition：當 power 不夠時，playing face down 都是 cost -1，所以當然要能夠獲得越多的 power 越好，所以使用 two pointer 從尾巴開始。

class Solution {
public:
    int bagOfTokensScore(vector<int>& tokens, int power) {
        sort(tokens.begin(), tokens.end());
 
        int n = tokens.size();
        int score = 0, res = 0;
 
        int i = 0, j = n - 1;
 
        while(i <= j) {
            if(power >= tokens[i]) {
                power -= tokens[i++];
                res = max(res, ++score);
            } else if(score > 0) {
                score--;
                power += tokens[j--];
            } else {
                break;
            }
        }
 
        return res;
    }
 
};

Source

• Bag of Tokens - LeetCode