Q - Given an m x n binary matrix mat, return the number of special positions in mat. A position (i, j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).
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class Solution {
public int numSpecial(int[][] mat) {
int count = 0;
int lenRow = mat.length; // Get the number of rows in the matrix
int lenCol = mat[0].length; // Get the number of columns in the matrix
// Iterate through each cell in the matrix
for (int i = 0; i < lenRow; i++) { // i = row
for (int j = 0; j < lenCol; j++) { // j = column
if (mat[i][j] == 1) { // Check if the current cell contains 1 (special element)
int spCol = 0;
int spRow = 0;
// Check the number of 1s in the same column as the current cell
for (int k = 0; k < lenRow; k++) {
if (mat[k][j] == 1) {
spCol++;
}
}
// Check the number of 1s in the same row as the current cell
for (int p = 0; p < lenCol; p++) {
if (mat[i][p] == 1) {
spRow++;
}
}
// If the cell is the sole '1' in its row and column, count it as special.
if ((spCol == 1) && (spRow == 1)) {
count++;
}
}
}
}
return count; // Return the count of special elements
}
}