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Problem solution · C++

Largest Magic Square

Largest Magic Square: 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
63 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 Largest Magic Square, 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

  • 63 lines of C++ from the credited upstream file 1895.cpp.
  • The implementation visibly relies on sequence storage.
  • 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 codeLargest Magic Square · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int largestMagicSquare(vector<vector<int>>& grid) {    const int m = grid.size();    const int n = grid[0].size();    // prefixRow[i][j] := the sum of the first j numbers in the i-th row    vector<vector<int>> prefixRow(m, vector<int>(n + 1));    // prefixCol[i][j] := the sum of the first j numbers in the i-th column    vector<vector<int>> prefixCol(n, vector<int>(m + 1));     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j) {        prefixRow[i][j + 1] = prefixRow[i][j] + grid[i][j];        prefixCol[j][i + 1] = prefixCol[j][i] + grid[i][j];      }     for (int k = min(m, n); k >= 2; --k)      if (containsMagicSquare(grid, prefixRow, prefixCol, k))        return k;     return 1;  }  private:  // Returns true if the grid contains any magic square of size k x k.  bool containsMagicSquare(const vector<vector<int>>& grid,                           const vector<vector<int>>& prefixRow,                           const vector<vector<int>>& prefixCol, int k) {    for (int i = 0; i + k - 1 < grid.size(); ++i)      for (int j = 0; j + k - 1 < grid[0].size(); ++j)        if (isMagicSquare(grid, prefixRow, prefixCol, i, j, k))          return true;    return false;  }   // Returns true if grid[i..i + k)[j..j + k) is a magic square.  bool isMagicSquare(const vector<vector<int>>& grid,                     const vector<vector<int>>& prefixRow,                     const vector<vector<int>>& prefixCol, int i, int j,                     int k) {    int diag = 0;    int antiDiag = 0;    for (int d = 0; d < k; ++d) {      diag += grid[i + d][j + d];      antiDiag += grid[i + d][j + k - 1 - d];    }    if (diag != antiDiag)      return false;    for (int d = 0; d < k; ++d) {      if (getSum(prefixRow, i + d, j, j + k - 1) != diag)        return false;      if (getSum(prefixCol, j + d, i, i + k - 1) != diag)        return false;    }    return true;  }   // Returns sum(grid[i][l..r]) or sum(grid[l..r][i]).  int getSum(const vector<vector<int>>& prefix, int i, int l, int r) {    return prefix[i][r + 1] - prefix[i][l];  }};

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