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

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