2530. Maximal Score After Applying K Operations

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problem

solution

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class Solution {
public:
    long long maxKelements(vector<int>& nums, int k) {
        long long ret = 0 ;
        // select max value in vector
        priority_queue <int> pq;
        for(int n:nums) pq.push(n);
        while(k--){            
            double t =pq.top();
            ret+=t;
            pq.pop();
            pq.push(ceil(t/3));
        }
        return ret;
        
    }
};

analysis

  • time complexity O(Klogn) , K is the number of query
  • space complexity O(n)