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1895. Largest Magic Square

Description

A k x k magic square is a k x k grid filled with integers such that every row sum, every column sum, and both diagonal sums are all equal. The integers in the magic square do not have to be distinct. Every 1 x 1 grid is trivially a magic square.

Given an m x n integer grid, return the size (i.e., the side length k) of the largest magic square that can be found within this grid.

 

Example 1:

Input: grid = [[7,1,4,5,6],[2,5,1,6,4],[1,5,4,3,2],[1,2,7,3,4]]
Output: 3
Explanation: The largest magic square has a size of 3.
Every row sum, column sum, and diagonal sum of this magic square is equal to 12.
- Row sums: 5+1+6 = 5+4+3 = 2+7+3 = 12
- Column sums: 5+5+2 = 1+4+7 = 6+3+3 = 12
- Diagonal sums: 5+4+3 = 6+4+2 = 12

Example 2:

Input: grid = [[5,1,3,1],[9,3,3,1],[1,3,3,8]]
Output: 2

 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• 1 <= grid[i][j] <= 106

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      private int[][] rowsum;
      private int[][] colsum;
  
      public int largestMagicSquare(int[][] grid) {
          int m = grid.length, n = grid[0].length;
          rowsum = new int[m + 1][n + 1];
          colsum = new int[m + 1][n + 1];
          for (int i = 1; i <= m; ++i) {
              for (int j = 1; j <= n; ++j) {
                  rowsum[i][j] = rowsum[i][j - 1] + grid[i - 1][j - 1];
                  colsum[i][j] = colsum[i - 1][j] + grid[i - 1][j - 1];
              }
          }
          for (int k = Math.min(m, n); k > 1; --k) {
              for (int i = 0; i + k - 1 < m; ++i) {
                  for (int j = 0; j + k - 1 < n; ++j) {
                      int i2 = i + k - 1, j2 = j + k - 1;
                      if (check(grid, i, j, i2, j2)) {
                          return k;
                      }
                  }
              }
          }
          return 1;
      }
  
      private boolean check(int[][] grid, int x1, int y1, int x2, int y2) {
          int val = rowsum[x1 + 1][y2 + 1] - rowsum[x1 + 1][y1];
          for (int i = x1 + 1; i <= x2; ++i) {
              if (rowsum[i + 1][y2 + 1] - rowsum[i + 1][y1] != val) {
                  return false;
              }
          }
          for (int j = y1; j <= y2; ++j) {
              if (colsum[x2 + 1][j + 1] - colsum[x1][j + 1] != val) {
                  return false;
              }
          }
          int s = 0;
          for (int i = x1, j = y1; i <= x2; ++i, ++j) {
              s += grid[i][j];
          }
          if (s != val) {
              return false;
          }
          s = 0;
          for (int i = x1, j = y2; i <= x2; ++i, --j) {
              s += grid[i][j];
          }
          if (s != val) {
              return false;
          }
          return true;
      }
  }
  

• class Solution {
  public:
      int largestMagicSquare(vector<vector<int>>& grid) {
          int m = grid.size(), n = grid.size();
          vector<vector<int>> rowsum(m + 1, vector<int>(n + 1));
          vector<vector<int>> colsum(m + 1, vector<int>(n + 1));
          for (int i = 1; i <= m; ++i) {
              for (int j = 1; j <= n; ++j) {
                  rowsum[i][j] = rowsum[i][j - 1] + grid[i - 1][j - 1];
                  colsum[i][j] = colsum[i - 1][j] + grid[i - 1][j - 1];
              }
          }
          for (int k = min(m, n); k > 1; --k) {
              for (int i = 0; i + k - 1 < m; ++i) {
                  for (int j = 0; j + k - 1 < n; ++j) {
                      int i2 = i + k - 1, j2 = j + k - 1;
                      if (check(grid, rowsum, colsum, i, j, i2, j2))
                          return k;
                  }
              }
          }
          return 1;
      }
  
      bool check(vector<vector<int>>& grid, vector<vector<int>>& rowsum, vector<vector<int>>& colsum, int x1, int y1, int x2, int y2) {
          int val = rowsum[x1 + 1][y2 + 1] - rowsum[x1 + 1][y1];
          for (int i = x1 + 1; i <= x2; ++i)
              if (rowsum[i + 1][y2 + 1] - rowsum[i + 1][y1] != val)
                  return false;
          for (int j = y1; j <= y2; ++j)
              if (colsum[x2 + 1][j + 1] - colsum[x1][j + 1] != val)
                  return false;
          int s = 0;
          for (int i = x1, j = y1; i <= x2; ++i, ++j)
              s += grid[i][j];
          if (s != val)
              return false;
          s = 0;
          for (int i = x1, j = y2; i <= x2; ++i, --j)
              s += grid[i][j];
          if (s != val)
              return false;
          return true;
      }
  };
  

• class Solution:
      def largestMagicSquare(self, grid: List[List[int]]) -> int:
          m, n = len(grid), len(grid[0])
          rowsum = [[0] * (n + 1) for _ in range(m + 1)]
          colsum = [[0] * (n + 1) for _ in range(m + 1)]
          for i in range(1, m + 1):
              for j in range(1, n + 1):
                  rowsum[i][j] = rowsum[i][j - 1] + grid[i - 1][j - 1]
                  colsum[i][j] = colsum[i - 1][j] + grid[i - 1][j - 1]
  
          def check(x1, y1, x2, y2):
              val = rowsum[x1 + 1][y2 + 1] - rowsum[x1 + 1][y1]
              for i in range(x1 + 1, x2 + 1):
                  if rowsum[i + 1][y2 + 1] - rowsum[i + 1][y1] != val:
                      return False
              for j in range(y1, y2 + 1):
                  if colsum[x2 + 1][j + 1] - colsum[x1][j + 1] != val:
                      return False
              s, i, j = 0, x1, y1
              while i <= x2:
                  s += grid[i][j]
                  i += 1
                  j += 1
              if s != val:
                  return False
              s, i, j = 0, x1, y2
              while i <= x2:
                  s += grid[i][j]
                  i += 1
                  j -= 1
              if s != val:
                  return False
              return True
  
          for k in range(min(m, n), 1, -1):
              i = 0
              while i + k - 1 < m:
                  j = 0
                  while j + k - 1 < n:
                      i2, j2 = i + k - 1, j + k - 1
                      if check(i, j, i2, j2):
                          return k
                      j += 1
                  i += 1
          return 1
  
  

• func largestMagicSquare(grid [][]int) int {
  	m, n := len(grid), len(grid[0])
  	rowsum := make([][]int, m+1)
  	colsum := make([][]int, m+1)
  	for i := 0; i <= m; i++ {
  		rowsum[i] = make([]int, n+1)
  		colsum[i] = make([]int, n+1)
  	}
  	for i := 1; i < m+1; i++ {
  		for j := 1; j < n+1; j++ {
  			rowsum[i][j] = rowsum[i][j-1] + grid[i-1][j-1]
  			colsum[i][j] = colsum[i-1][j] + grid[i-1][j-1]
  		}
  	}
  	for k := min(m, n); k > 1; k-- {
  		for i := 0; i+k-1 < m; i++ {
  			for j := 0; j+k-1 < n; j++ {
  				i2, j2 := i+k-1, j+k-1
  				if check(grid, rowsum, colsum, i, j, i2, j2) {
  					return k
  				}
  			}
  		}
  	}
  	return 1
  }
  
  func check(grid, rowsum, colsum [][]int, x1, y1, x2, y2 int) bool {
  	val := rowsum[x1+1][y2+1] - rowsum[x1+1][y1]
  	for i := x1 + 1; i < x2+1; i++ {
  		if rowsum[i+1][y2+1]-rowsum[i+1][y1] != val {
  			return false
  		}
  	}
  	for j := y1; j < y2+1; j++ {
  		if colsum[x2+1][j+1]-colsum[x1][j+1] != val {
  			return false
  		}
  	}
  	s := 0
  	for i, j := x1, y1; i <= x2; i, j = i+1, j+1 {
  		s += grid[i][j]
  	}
  	if s != val {
  		return false
  	}
  	s = 0
  	for i, j := x1, y2; i <= x2; i, j = i+1, j-1 {
  		s += grid[i][j]
  	}
  	if s != val {
  		return false
  	}
  	return true
  }
  

• function largestMagicSquare(grid: number[][]): number {
      let m = grid.length,
          n = grid[0].length;
      // 前缀和
      let rowSum = Array.from({ length: m + 1 }, (v, i) => new Array(n + 1).fill(0)),
          colSum = Array.from({ length: m + 1 }, v => new Array(n + 1).fill(0));
      for (let i = 0; i < m; i++) {
          rowSum[i + 1][1] = grid[i][0];
          for (let j = 1; j < n; j++) {
              rowSum[i + 1][j + 1] = rowSum[i + 1][j] + grid[i][j];
          }
      }
  
      for (let j = 0; j < n; j++) {
          colSum[1][j + 1] = grid[0][j];
          for (let i = 1; i < m; i++) {
              colSum[i + 1][j + 1] = colSum[i][j + 1] + grid[i][j];
          }
      }
      // console.log(rowSum, colSum)
      // 寻找最大k
      for (let k = Math.min(m, n); k > 1; k--) {
          for (let i = 0; i + k - 1 < m; i++) {
              for (let j = 0; j + k - 1 < n; j++) {
                  let x2 = i + k - 1,
                      y2 = j + k - 1;
                  if (valid(grid, rowSum, colSum, i, j, x2, y2)) {
                      return k;
                  }
              }
          }
      }
      return 1;
  }
  
  function valid(
      grid: number[][],
      rowSum: number[][],
      colSum: number[][],
      x1: number,
      y1: number,
      x2: number,
      y2: number,
  ): boolean {
      let diff = rowSum[x1 + 1][y2 + 1] - rowSum[x1 + 1][y1];
      // 行
      for (let i = x1 + 1; i <= x2; i++) {
          if (diff != rowSum[i + 1][y2 + 1] - rowSum[i + 1][y1]) {
              return false;
          }
      }
      // 列
      for (let j = y1; j <= y2; j++) {
          if (diff != colSum[x2 + 1][j + 1] - colSum[x1][j + 1]) {
              return false;
          }
      }
      // 主队对角线
      let mainSum = diff;
      for (let i = x1, j = y1; i <= x2; i++, j++) {
          mainSum -= grid[i][j];
      }
      if (mainSum != 0) return false;
      // 副对角线
      let subSum = diff;
      for (let i = x1, j = y2; i <= x2; i++, j--) {
          subSum -= grid[i][j];
      }
      if (subSum != 0) return false;
      return true;
  }
  
  

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