EASY SOLUTION USING MAX HEAP || C++
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int miceAndCheese(vector<int>& reward1, vector<int>& reward2, int k) {\n        int ans=0;      \n        int n=reward1.size();\n        priority_queue<pair<pair<int , int> , int>>pq;  \n\n        for(int i=0;i<n;i++)            \n            pq.push({{reward1[i]-reward2[i],reward1[i]},reward2[i]});\n\n        while(!pq.empty()&&k--){\n            ans=ans+pq.top().first.second;\n            pq.pop();\n        }   \n\n        while(!pq.empty()){\n            ans=ans+pq.top().second;\n            pq.pop();\n        }        \n        return ans;\n    }\n};\n```
['Array', 'Heap (Priority Queue)', 'C++']