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This approach involves iterating through each element of the matrix and checking if it is '1'. For each such element, all other elements in its row and column are checked to ensure they are '0'. This approach is straightforward but can be optimized.
Time Complexity: O(m * n * (m + n)), where m is the number of rows and n is the number of columns.
Space Complexity: O(1), since we use a constant amount of space.
1public class Solution {
2 public int NumSpecial(int[][] mat) {
3 int m = mat.Length;
4 int n = mat[0].Length;
5 int specialCount = 0;
6 for (int i = 0; i < m; i++) {
7 for (int j = 0; j < n; j++) {
8 if (mat[i][j] == 1) {
9 bool isSpecial = true;
10 for (int k = 0; k < m; k++) {
11 if (k != i && mat[k][j] == 1) {
12 isSpecial = false;
13 break;
14 }
15 }
16 for (int k = 0; k < n; k++) {
17 if (k != j && mat[i][k] == 1) {
18 isSpecial = false;
19 break;
20 }
21 }
22 if (isSpecial) specialCount++;
23 }
24 }
25 }
26 return specialCount;
27 }
28}The C# solution applies a simple nested loop iteration strategy to assess each matrix element. For '1's found, both the respective row and column are scanned to determine if the element satisfies the special condition.
In this approach, two auxiliary arrays are used to track the number of '1's in each row and column. The matrix is then iterated to find the positions where both the row and column contain exactly one '1'. This approach reduces the need for repeatedly checking rows and columns.
Time Complexity: O(m * n), because we only iterate through the matrix twice.
Space Complexity: O(m + n), for storing row and column counts.
1
This JavaScript function builds arrays to record the count of '1's for each row and column. It then checks for special positions using these counts.