{ "data": { "question": { "questionId": "985", "questionFrontendId": "948", "categoryTitle": "Algorithms", "boundTopicId": 2420, "title": "Bag of Tokens", "titleSlug": "bag-of-tokens", "content": "
You have an initial power of power
, an initial score of 0
, and a bag of tokens
where tokens[i]
is the value of the ith
token (0-indexed).
Your goal is to maximize your total score by potentially playing each token in one of two ways:
\n\ntokens[i]
, you may play the ith
token face up, losing tokens[i]
power and gaining 1
score.1
, you may play the ith
token face down, gaining tokens[i]
power and losing 1
score.Each token may be played at most once and in any order. You do not have to play all the tokens.
\n\nReturn the largest possible score you can achieve after playing any number of tokens.
\n\n\n
Example 1:
\n\n\nInput: tokens = [100], power = 50\nOutput: 0\nExplanation: Playing the only token in the bag is impossible because you either have too little power or too little score.\n\n\n
Example 2:
\n\n\nInput: tokens = [100,200], power = 150\nOutput: 1\nExplanation: Play the 0th token (100) face up, your power becomes 50 and score becomes 1.\nThere is no need to play the 1st token since you cannot play it face up to add to your score.\n\n\n
Example 3:
\n\n\nInput: tokens = [100,200,300,400], power = 200\nOutput: 2\nExplanation: Play the tokens in this order to get a score of 2:\n1. Play the 0th token (100) face up, your power becomes 100 and score becomes 1.\n2. Play the 3rd token (400) face down, your power becomes 500 and score becomes 0.\n3. Play the 1st token (200) face up, your power becomes 300 and score becomes 1.\n4. Play the 2nd token (300) face up, your power becomes 0 and score becomes 2.\n\n\n
\n
Constraints:
\n\n0 <= tokens.length <= 1000
0 <= tokens[i], power < 104
你的初始 能量 为 P
,初始 分数 为 0
,只有一包令牌 tokens
。其中 tokens[i]
是第 i
个令牌的值(下标从 0 开始)。
令牌可能的两种使用方法如下:
\n\ntoken[i]
点 能量 ,可以将令牌 i
置为正面朝上,失去 token[i]
点 能量 ,并得到 1
分 。1
分 ,可以将令牌 i
置为反面朝上,获得 token[i]
点 能量 ,并失去 1
分 。每个令牌 最多 只能使用一次,使用 顺序不限 ,不需 使用所有令牌。
\n\n在使用任意数量的令牌后,返回我们可以得到的最大 分数 。
\n\n\n\n
示例 1:
\n\n\n输入:tokens = [100], P = 50\n输出:0\n解释:无法使用唯一的令牌,因为能量和分数都太少了。\n\n
示例 2:
\n\n\n输入:tokens = [100,200], P = 150\n输出:1\n解释:令牌 0 正面朝上,能量变为 50,分数变为 1 。不必使用令牌 1 ,因为你无法使用它来提高分数。\n\n
示例 3:
\n\n\n输入:tokens = [100,200,300,400], P = 200\n输出:2\n解释:按下面顺序使用令牌可以得到 2 分:\n1. 令牌 0 正面朝上,能量变为 100 ,分数变为 1\n2. 令牌 3 正面朝下,能量变为 500 ,分数变为 0\n3. 令牌 1 正面朝上,能量变为 300 ,分数变为 1\n4. 令牌 2 正面朝上,能量变为 0 ,分数变为 2\n\n
\n\n
提示:
\n\n0 <= tokens.length <= 1000
0 <= tokens[i], P < 104
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