Beginner Friendly - Hinglish-Comments-Intuition- Approach -easy to understand 
Intuition
Humare paas classes hain, jisme pass aur total students ka ratio diya hai.
Hume extraStudents dene hain taaki pass ratio ko maximize kiya ja sake.
Har
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Humare paas classes hain, jisme pass aur total students ka ratio diya hai.\n2. Hume extraStudents dene hain taaki pass ratio ko maximize kiya ja sake.\n3. Har class ke liye hum ek delta (improvement) calculate karte hain, jo batata hai ki ek extra student add karne se ratio me kitna improvement hoga.\n4. Hum priority queue (max heap) ka use karte hain taaki har baar wo class select ho jisme sabse zyada improvement ho.\n5. Fir extraStudents ko classes par apply karte hain, aur finally average pass ratio nikalte hain.\n\n# Step-by-Step:\nDelta Calculation: Har class ka delta (improvement) calculate karo.\nPriority Queue: Delta ke basis par classes ko queue me store karo.\nApply Extra Students: Har step me top class ko select kar ke uske pass aur total ko increment karo.\nFinal Average: Sab classes ka updated pass/total ratio ka average calculate karo.\nDry Run Example:\nGiven:\nclasses = [[1, 2], [3, 5], [2, 2]]\nextraStudents = 2\nInitial delta calculation:\n\nClass 1: delta = 0.1667    // ye wala max heap me top pr aayega \nClass 2: delta = 0.0667\nClass 3: delta = 0\nApply extra students:\n\nApply to Class 1 (max delta) \u2192 becomes (2 pass, 3 total).\nApply again to Class 1 \u2192 becomes (3 pass, 4 total).\nFinal average:\n\nClass 1: 3/4 = 0.75\nClass 2: 3/5 = 0.6\nClass 3: 2/2 = 1\nFinal average = (0.75 + 0.6 + 1) / 3 = 0.7833.\n\nResult:\nAverage pass ratio = 0.7833\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   O(n log n).\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n      O(n)\n# Code\n```cpp []\nclass Solution {\npublic:\n   // Function to pass ratio improvement cal. krne k liye \ndouble getDelta(int pass, int total) {\n    return (double)(pass + 1) / (total + 1) - (double)pass / total;\n}\n\ndouble maxAverageRatio(vector<vector<int>>& classes, int k) {\n    priority_queue<pair<double, pair<int, int>>> pq;\n\n          int pass , total ;\n    for(int i =0 ; i<classes.size();i++){\n            int pass = classes[i][0];\n            int total = classes[i][1];\n    // yha pr sab me 1 add krke unka increasing ratio kitna hai vo aur pair store kr diye \n           pq.push({getDelta(pass,total),{pass,total}}) ;\n    }\n\n    // abb assign krenge extra wale ko \n    while(k--){\n         // top wal me increment jyda hoga iske pass and total value ko increse krenge \n         int pass = pq.top().second.first ; \n         int total  =  pq.top().second.second ; \n         pass++ ;\n         total++ ; \n         // top ko remove krke new increment ratio wala value re-insert krenge \n         pq.pop();\n         pq.push({getDelta(pass,total),{pass,total}}) ; \n\n    }\n\n             double avg = 0.0 ; \n    while(!pq.empty()){\n        int pass = pq.top().second.first ; \n         int total  =  pq.top().second.second ; \n         avg += (double)pass / total ;    // double value k liye type caste kiye hai \n         pq.pop() ;  \n    }\nreturn avg/classes.size() ; \n}\n};\n```
['Greedy', 'Heap (Priority Queue)', 'C++']