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This approach involves counting the number of ones and zeros in each row and each column beforehand. Then, for each element in the grid, compute the corresponding value in the difference matrix using the precomputed counts.
Time Complexity: O(m * n) where m is the number of rows and n is the number of columns. This involves going through each element of the grid twice; once for counting and once for filling the difference matrix.
Space Complexity: O(m + n) for storing counts of ones and zeros for each row and column.
1def difference_matrix(grid):
2 m, n = len(grid), len(grid[0])
3 ones_row = [0] * m
4 zeros_row = [0] * m
5 ones_col = [0] * n
6 zeros_col = [0] * n
7
8 # Precompute the number of ones and zeros in each row and column
9 for i in range(m):
10 for j in range(n):
11 if grid[i][j] == 1:
12 ones_row[i] += 1
13 ones_col[j] += 1
14 else:
15 zeros_row[i] += 1
16 zeros_col[j] += 1
17
18 diff = [[0] * n for _ in range(m)]
19 # Create the difference matrix
20 for i in range(m):
21 for j in range(n):
22 diff[i][j] = ones_row[i] + ones_col[j] - zeros_row[i] - zeros_col[j]
23 return diff
24The Python solution effectively uses lists to precompute the number of ones and zeros. The difference matrix is then calculated for each element in the grid leveraging these precomputed values, aligning with a straightforward but efficient approach.
This approach seeks to optimize the space complexity by calculating necessary counts and the result matrix in a single pass, thereby avoiding separate storage for ones and zeros counts.
Time Complexity: O(m * n) because every element is iterated over during the counting and result generation stages.
Space Complexity: O(n) due to the storage of column ones, improving from O(m + n).
using System.Linq;
public class Solution {
public int[][] DifferenceMatrixSinglePass(int[][] grid) {
int m = grid.Length;
int n = grid[0].Length;
int[][] diff = new int[m][];
int[] onesCol = new int[n];
// Calculate the number of ones in each column
for (int j = 0; j < n; j++) {
onesCol[j] = grid.Sum(row => row[j]);
}
for (int i = 0; i < m; i++) {
diff[i] = new int[n];
int onesRow = grid[i].Sum();
int zerosRow = n - onesRow;
for (int j = 0; j < n; j++) {
int zerosCol = m - onesCol[j];
diff[i][j] = onesRow + onesCol[j] - zerosRow - zerosCol;
}
}
return diff;
}
}
This C# solution mirrors the efficient, liner single-pass calculation methodology, leveraging LINQ for sum operations to alleviate verbosity and streamline the algorithm, preserving both evolved logic and resource demands.