Cinema Seat Allocation LeetCode solution in C++

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class Solution {
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 public:
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  int maxNumberOfFamilies(int n, vector<vector<int>> &reservedSeats) {
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    unordered_map<int, vector<int>> o;
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    for (auto r : reservedSeats) {
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      o[r[0]].push_back(r[1]);
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    }
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    int ans = 2 * (n - o.size()); // seats with no occupancy directly adding 2
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    // if we iterate from 1 to n (10 ^ 9) we cross time limit
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    for (auto p : o) {
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      // get reserved seats
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      vector<int> seats = p.second;
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      int occupied_mask = 0;
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      // representing occupied seats as bits
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      for (int s : seats) {
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        if (s > 1 && s < 10) {
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          occupied_mask += (int)pow(2, 10 - s);
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        }
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      }
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      // checking 3 seating configurations
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      int masks[3] = {480, 120, 30};
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      int row_ans = 0;
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      bool res1 = (masks[0] & occupied_mask) == 0;
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      bool res2 = (masks[1] & occupied_mask) == 0;
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      bool res3 = (masks[2] & occupied_mask) == 0;
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      // if all 3 configurations are empty, 2 families
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      // if either one of them are empty, 1 family
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      if (res1 && res2 && res3)
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        row_ans += 2;
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      else if (res1 || res2 || res3)
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        row_ans += 1;
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      ans += row_ans;
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    }
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    return ans;
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  }
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};