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Problem solution · Java

Largest Magic Square

Largest Magic Square: a Java 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
57 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

  • 57 lines of Java from the credited upstream file 1895.java.
  • 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 · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int largestMagicSquare(int[][] grid) {    final int m = grid.length;    final int n = grid[0].length;    // prefixRow[i][j] := the sum of the first j numbers in the i-th row    int[][] prefixRow = new int[m][n + 1];    // prefixCol[i][j] := the sum of the first j numbers in the i-th column    int[][] prefixCol = new int[n][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 = Math.min(m, n); k >= 2; --k)      if (containsMagicSquare(grid, prefixRow, prefixCol, k))        return k;     return 1;  }   // Returns true if the grid contains any magic square of size k x k.  private boolean containsMagicSquare(int[][] grid, int[][] prefixRow, int[][] prefixCol, int k) {    for (int i = 0; i + k - 1 < grid.length; ++i)      for (int j = 0; j + k - 1 < grid[0].length; ++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.  private boolean isMagicSquare(int[][] grid, int[][] prefixRow, 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]).  private int getSum(int[][] prefix, int i, int l, int r) {    return prefix[i][r + 1] - prefix[i][l];  }}

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