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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + double maxAverageRatio(vector<vector<int>>& classes, int extraStudents) { |
| 4 | + int classesLen = classes.size(); |
| 5 | + |
| 6 | + // pq to know which class will increase the avg the most |
| 7 | + priority_queue<pair<double, int>> valueQueue; |
| 8 | + double finalAvg = 0; |
| 9 | + double localAvg = 0; |
| 10 | + |
| 11 | + |
| 12 | + // going through all classes adding them to finalAvg and |
| 13 | + // placing their avg increase into the pq if we were to add |
| 14 | + // an extra student there |
| 15 | + for (int i = 0; i < classesLen; ++i) { |
| 16 | + localAvg = static_cast<double>(classes[i][0]) / classes[i][1]; |
| 17 | + valueQueue.push({(static_cast<double>(classes[i][0] + 1) / (classes[i][1] + 1)) - localAvg, i}); |
| 18 | + finalAvg += localAvg; |
| 19 | + } |
| 20 | + |
| 21 | + // devide the avg once after the loop |
| 22 | + finalAvg /= classesLen; |
| 23 | + |
| 24 | + |
| 25 | + // while there are students to distribute we go through the pq |
| 26 | + while (extraStudents > 0) { |
| 27 | + // we take the class with highest avg inc value |
| 28 | + auto top = valueQueue.top(); |
| 29 | + valueQueue.pop(); |
| 30 | + |
| 31 | + // add the value / classLen to the finalAvg |
| 32 | + finalAvg += top.first / classesLen; |
| 33 | + |
| 34 | + // we add a student to that class |
| 35 | + ++classes[top.second][1]; |
| 36 | + ++classes[top.second][0]; |
| 37 | + |
| 38 | + // recalculate that class avg inc if we were to add an |
| 39 | + // extra student |
| 40 | + localAvg = static_cast<double>(classes[top.second][0]) / classes[top.second][1]; |
| 41 | + valueQueue.push({(static_cast<double>(classes[top.second][0] + 1) / (classes[top.second][1] + 1)) - localAvg, top.second}); |
| 42 | + |
| 43 | + // decrement the extraStudents left |
| 44 | + --extraStudents; |
| 45 | + } |
| 46 | + |
| 47 | + return finalAvg; |
| 48 | + } |
| 49 | +}; |
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