description: On an endless checkered sheet of paper, n cells are chosen and colored in three colors, where n is divisible by 3. It turns out that there are exactly n/3 marked cells of each of three colors! 
  
  Find the largest such k that it's possible to choose k/3 cells of each color, remove all other marked cells, and then select three rectangles with sides parallel to the grid lines so that the following conditions hold:
  
    * No two rectangles can intersect (but they can share a part of the boundary). In other words, the area of intersection of any two of these rectangles must be 0.
    * The i-th rectangle contains all the chosen cells of the i-th color and no chosen cells of other colors, for i = 1, 2, 3. 
  
  Input
  
  The first line of the input contains a single integer n — the number of the marked cells (3 ≤ n ≤ 10^5, n is divisible by 3).
  
  The i-th of the following n lines contains three integers x_i, y_i, c_i (|x_i|,|y_i| ≤ 10^9; 1 ≤ c_i ≤ 3), where (x_i, y_i) are the coordinates of the i-th marked cell and c_i is its color.
  
  It's guaranteed that all cells (x_i, y_i) in the input are distinct, and that there are exactly n/3 cells of each color.
  
  Output
  
  Output a single integer k — the largest number of cells you can leave.
  
  Examples
  
  Input
  
  
  9
  2 3 1
  4 1 2
  2 1 3
  3 4 1
  5 3 2
  4 4 3
  2 4 1
  5 2 2
  3 5 3
  
  
  Output
  
  
  6
  
  
  Input
  
  
  3
  1 1 1
  2 2 2
  3 3 3
  
  
  Output
  
  
  3
  
  Note
  
  In the first sample, it's possible to leave 6 cells with indexes 1, 5, 6, 7, 8, 9.
  
  In the second sample, it's possible to leave 3 cells with indexes 1, 2, 3.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int inf = (int)1e9 + 1;
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    vector<pair<int, int>> v[3][2];
    vector<vector<int>> perms = {{0, 1, 2}, {0, 2, 1}, {1, 0, 2},
                                 {1, 2, 0}, {2, 0, 1}, {2, 1, 0}};
    for (int i = 0; i < n; i++) {
      int x, y, c;
      cin >> x >> y >> c;
      c--;
      v[c][0].emplace_back(make_pair(x, y));
      v[c][1].emplace_back(make_pair(x, y));
    }
    for (auto& i : v) {
      sort(i[0].begin(), i[0].end());
      sort(i[1].begin(), i[1].end(), [&](pair<int, int>& a, pair<int, int>& b) {
        return make_pair(a.second, a.first) < make_pair(b.second, b.first);
      });
    }
    auto check = [&](int k) -> bool {
      bool ret = false;
      bool flg;
      int x_lim, y_lim;
      for (const auto& p : perms) {
        x_lim = -inf;
        flg = true;
        for (auto x : p) {
          int ths = 0;
          int n_x = x_lim;
          for (auto ele : v[x][0]) {
            if (ths == k) {
              break;
            }
            if (ele.first > x_lim) {
              n_x = max(n_x, ele.first);
              ++ths;
            }
          }
          if (ths < k) {
            flg = false;
          }
          x_lim = n_x;
        }
        ret |= flg;
      }
      for (const auto& p : perms) {
        y_lim = -inf;
        flg = true;
        for (auto x : p) {
          int ths = 0;
          int n_y = y_lim;
          for (auto ele : v[x][1]) {
            if (ths == k) {
              break;
            }
            if (ele.second > y_lim) {
              n_y = max(n_y, ele.second);
              ++ths;
            }
          }
          if (ths < k) {
            flg = false;
          }
          y_lim = n_y;
        }
        ret |= flg;
      }
      for (const auto& p : perms) {
        flg = true;
        x_lim = v[p[0]][0][k - 1].first;
        int ths = 0;
        for (auto ele : v[p[1]][1]) {
          if (ths == k) {
            break;
          }
          if (ele.first > x_lim) {
            y_lim = ele.second;
            ++ths;
          }
        }
        if (ths < k) {
          flg = false;
        }
        ths = 0;
        for (auto ele : v[p[2]][0]) {
          ths += (ele.first > x_lim && ele.second > y_lim);
        }
        if (ths < k) {
          flg = false;
        }
        ret |= flg;
      }
      for (const auto& p : perms) {
        flg = true;
        y_lim = v[p[0]][1][k - 1].second;
        int ths = 0;
        for (auto ele : v[p[1]][0]) {
          if (ths == k) {
            break;
          }
          if (ele.second > y_lim) {
            x_lim = ele.first;
            ++ths;
          }
        }
        if (ths < k) {
          flg = false;
        }
        ths = 0;
        for (auto ele : v[p[2]][0]) {
          ths += (ele.first > x_lim && ele.second > y_lim);
        }
        if (ths < k) {
          flg = false;
        }
        ret |= flg;
      }
      for (const auto& p : perms) {
        flg = true;
        x_lim = v[p[0]][0][n / 3 - k].first;
        int ths = 0;
        for (auto ele : v[p[1]][1]) {
          if (ths == k) {
            break;
          }
          if (ele.first < x_lim) {
            y_lim = ele.second;
            ++ths;
          }
        }
        if (ths < k) {
          flg = false;
        }
        ths = 0;
        for (auto ele : v[p[2]][0]) {
          ths += (ele.first < x_lim && ele.second > y_lim);
        }
        if (ths < k) {
          flg = false;
        }
        ret |= flg;
      }
      for (const auto& p : perms) {
        flg = true;
        y_lim = v[p[0]][1][n / 3 - k].second;
        int ths = 0;
        for (auto ele : v[p[1]][0]) {
          if (ths == k) {
            break;
          }
          if (ele.second < y_lim) {
            x_lim = ele.first;
            ++ths;
          }
        }
        if (ths < k) {
          flg = false;
        }
        ths = 0;
        for (auto ele : v[p[2]][0]) {
          ths += (ele.first > x_lim && ele.second < y_lim);
        }
        if (ths < k) {
          flg = false;
        }
        ret |= flg;
      }
      return ret;
    };
    int l = 1, r = n / 3;
    while (l < r) {
      int mid = (l + r + 1) / 2;
      if (check(mid)) {
        l = mid;
      } else {
        r = mid - 1;
      }
    }
    cout << 3 * l << '\n';
    return 0;
  }
  