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Problem solution · C++

Maximum Value Sum by Placing Three Rooks I

Maximum Value Sum by Placing Three Rooks I: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
64 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Maximum Value Sum by Placing Three Rooks I, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 64 lines of C++ from the credited upstream file 3256.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 7 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Value Sum by Placing Three Rooks I · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long maximumValueSum(vector<vector<int>>& board) {    const int m = board.size();    const int n = board[0].size();    long ans = LONG_MIN;    using T = tuple<long, int, int>;    vector<vector<T>> rows(m);  // [(val, i, j)]    vector<vector<T>> cols(n);  // [(val, i, j)]    set<T> rowSet;              // {(val, i, j)}    set<T> colSet;              // {(val, i, j)}    set<T> topNine;             // {(val, i, j)}     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j) {        rows[i].emplace_back(board[i][j], i, j);        cols[j].emplace_back(board[i][j], i, j);      }     auto getTop3 = [](vector<T>& row) -> vector<T> {      partial_sort(row.begin(),                   row.begin() + min(3, static_cast<int>(row.size())),                   row.end(), greater<>());      row.resize(min(3, (int)row.size()));      return row;    };     for (vector<T>& row : rows) {      row = getTop3(row);      rowSet.insert(row.begin(), row.end());    }     for (vector<T>& col : cols) {      col = getTop3(col);      colSet.insert(col.begin(), col.end());    }     set_intersection(rowSet.begin(), rowSet.end(), colSet.begin(), colSet.end(),                     inserter(topNine, topNine.begin()));     // At least 9 positions are required on the board to place 3 rooks such that    // none can attack another.    if (topNine.size() > 9) {      auto it = topNine.begin();      advance(it, topNine.size() - 9);      topNine.erase(topNine.begin(), it);    }     for (auto it1 = topNine.begin(); it1 != topNine.end(); ++it1)      for (auto it2 = next(it1); it2 != topNine.end(); ++it2)        for (auto it3 = next(it2); it3 != topNine.end(); ++it3) {          const auto [val1, i1, j1] = *it1;          const auto [val2, i2, j2] = *it2;          const auto [val3, i3, j3] = *it3;          if (i1 == i2 || i1 == i3 || i2 == i3 ||  //              j1 == j2 || j1 == j3 || j2 == j3)            continue;          ans = max(ans, val1 + val2 + val3);        }     return ans;  }};

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