Special Positions in a Binary Matrix LeetCode Problem

 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).

Example 1:



Input: mat = [[1,0,0],[0,0,1],[1,0,0]]

Output: 1

Explanation: (1, 2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:



Input: mat = [[1,0,0],[0,1,0],[0,0,1]]

Output: 3

Explanation: (0, 0), (1, 1) and (2, 2) are special positions.




class Solution {
    public int numSpecial(int[][] mat) {

        int row = mat.length; // Number of rows in the matrix
        int col = mat[0].length; // Number of columns in the matrix

        // Arrays to store the count of 1s in each row and column
        int[] row_count = new int[row]; // Stores count of 1s in each row
        int[] col_count = new int[col]; // Stores count of 1s in each column

        // First pass: Count the number of 1s in each row and column
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (mat[i][j] == 1) {
                    row_count[i]++; // Increase count of 1s in this row
                    col_count[j]++; // Increase count of 1s in this column
                }
            }
        }

        int special = 0; // To store the count of special positions

        // Second pass: Check for special positions
        for (int i = 0; i < row; i++) {
            if (row_count[i] == 1) { // If this row has exactly one '1'
                for (int j = 0; j < col; j++) {
                    if (col_count[j] == 1 && mat[i][j] == 1) { // If this column also has exactly one '1'
                        special++; // Found a special position
                    }
                }
            }
        }

        return special; // Return the count of special positions
    }
}

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