Bag of Tokens in C++

Suppose we have an initial power P, an initial score of 0 points, and one bag of tokens. Now each token can be used at most once, there is a value token[i], and has potentially two ways to use it, these are as follows −

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

We have to find the largest number of points that we can have after playing any number of tokens.

So if the input is like tokens = [100,200,300,400] and P = 200, then the output will be 2.

To solve this, we will follow these steps −

n := size of array v, ret := 0
sort the v array
set i := 0 and j := n – 1, curr :=
while i <= j and x >= v[i]
- while i <= j and x >= v[i]
  - decrease x by v[i], increase curr and i by 1
- ret := max of curr and ret
- while j >= i and curr is not 0 and x < v[i]
  - increase x by v[j], decrease curr and j by 1
return ret

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int bagOfTokensScore(vector<int>& v, int x) {
      int n = v.size();
      int ret = 0;
      sort(v.begin(), v.end());
      int i = 0;
      int j = n - 1;
      int curr = 0;
      while(i <= j && x >= v[i]){
         while(i <= j && x >= v[i]){
            x -= v[i];
            curr++;
            i++;
         }
         ret = max(curr, ret);
         while(j >= i && curr && x < v[i]){
            curr--;
            x += v[j];
            j--;
         }
      }
      return ret;
   }
};
main(){
   vector<int> v1 = {100,200,300,400};
   Solution ob;
   cout << (ob.bagOfTokensScore(v1, 200));
}