Sponsored
Sponsored
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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int** differenceMatrix(int** grid, int gridSize, int* gridColSize, int* returnSize, int** returnColumnSizes) {
5 int* onesRow = (int*)calloc(gridSize, sizeof(int));
6 int* zerosRow = (int*)calloc(gridSize, sizeof(int));
7 int* onesCol = (int*)calloc(gridColSize[0], sizeof(int));
8 int* zerosCol = (int*)calloc(gridColSize[0], sizeof(int));
9
10 // Precompute the number of ones and zeros in each row and column
11 for (int i = 0; i < gridSize; i++) {
12 for (int j = 0; j < gridColSize[i]; j++) {
13 if (grid[i][j] == 1) {
14 onesRow[i]++;
15 onesCol[j]++;
16 } else {
17 zerosRow[i]++;
18 zerosCol[j]++;
19 }
20 }
21 }
22
23 // Create the difference matrix
24 int** diff = (int**)malloc(gridSize * sizeof(int*));
25 *returnSize = gridSize;
26 *returnColumnSizes = (int*)malloc(gridSize * sizeof(int));
27
28 for (int i = 0; i < gridSize; i++) {
29 (*returnColumnSizes)[i] = gridColSize[i];
30 diff[i] = (int*)malloc(gridColSize[i] * sizeof(int));
31 for (int j = 0; j < gridColSize[i]; j++) {
32 diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j];
33 }
34 }
35
36 free(onesRow);
37 free(zerosRow);
38 free(onesCol);
39 free(zerosCol);
40 return diff;
41}
42In this C solution, we first initialize arrays to store the count of ones and zeros for each row and column. We iterate over the grid to fill these arrays. Then, we calculate the difference matrix using these precomputed values. Finally, the result is returned as a dynamically allocated 2D array.
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).
1
This JavaScript code optimizes calculation by keeping count computations in separate sections, using reduce for summing row ones, resulting in a less memory-taxing solution while retaining precision and promptness of result construction.