You have an initial power P, an initial score of 0 points, and a bag of tokens.

Each token can be used at most once, has a value token[i], and has potentially two ways to use it.

• If we have at least token[i] power, we may play the token face up, losing token[i] power, and gaining 1 point.
• If we have at least 1 point, we may play the token face down, gaining token[i]power, and losing 1 point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

1. tokens.length <= 1000
2. 0 <= tokens[i] < 10000
3. 0 <= P < 10000

Solution: Greedy + Two Pointers

Sort the tokens, gain points from the low end gain power from the high end.

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua, Running time: 8 ms
class Solution {
public:
  int bagOfTokensScore(vector<int>& tokens, int P) {
    sort(begin(tokens), end(tokens));
    int points = 0;
    int ans = 0;
    int i = 0;
    int j = tokens.size() - 1;
    
    while (i <= j)
      if (P >= tokens[i]) {
        P -= tokens[i++];
        ans = max(ans, ++points);        
      } else if (points > 0) {
        P += tokens[j--];
        --points;
      } else {
        break;
      }
    return ans;
  }
};

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