- Translate each rule into one explicit state update.
- Maintain the invariant after every processed item.
- Return the accumulated state once all relevant input has been handled.
Code notes
- 47 lines of Python from the credited upstream file 1895.py.
- The implementation visibly relies on sequence storage.
- No explicit 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.
Use this to learn the idea, then write your own version.
1class Solution:2 def largestMagicSquare(self, grid: list[list[int]]) -> int:3 m = len(grid)4 n = len(grid[0])5 6 prefixRow = [[0] * (n + 1) for _ in range(m)]7 8 prefixCol = [[0] * (m + 1) for _ in range(n)]9 10 for i in range(m):11 for j in range(n):12 prefixRow[i][j + 1] = prefixRow[i][j] + grid[i][j]13 prefixCol[j][i + 1] = prefixCol[j][i] + grid[i][j]14 15 def isMagicSquare(i: int, j: int, k: int) -> bool:16 """Returns True if grid[i..i + k)[j..j + k) is a magic square."""17 diag, antiDiag = 0, 018 for d in range(k):19 diag += grid[i + d][j + d]20 antiDiag += grid[i + d][j + k - 1 - d]21 if diag != antiDiag:22 return False23 for d in range(k):24 if self._getSum(prefixRow, i + d, j, j + k - 1) != diag:25 return False26 if self._getSum(prefixCol, j + d, i, i + k - 1) != diag:27 return False28 return True29 30 def containsMagicSquare(k: int) -> bool:31 """Returns True if the grid contains any magic square of size k x k."""32 for i in range(m - k + 1):33 for j in range(n - k + 1):34 if isMagicSquare(i, j, k):35 return True36 return False37 38 for k in range(min(m, n), 1, -1):39 if containsMagicSquare(k):40 return k41 42 return 143 44 def _getSum(self, prefix: list[list[int]], i: int, l: int, r: int) -> int:45 """Returns sum(grid[i][l..r]) or sum(grid[l..r][i])."""46 return prefix[i][r + 1] - prefix[i][l]47