Class Solution


  • public class Solution
    extends Object
    1582 - Special Positions in a Binary Matrix\. Easy 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** ). **Example 1:** ![](https://assets.leetcode.com/uploads/2021/12/23/special1.jpg) **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:** ![](https://assets.leetcode.com/uploads/2021/12/24/special-grid.jpg) **Input:** mat = \[\[1,0,0],[0,1,0],[0,0,1]] **Output:** 3 **Explanation:** (0, 0), (1, 1) and (2, 2) are special positions. **Constraints:** * `m == mat.length` * `n == mat[i].length` * `1 <= m, n <= 100` * `mat[i][j]` is either `0` or `1`.
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • numSpecial

        public int numSpecial​(int[][] mat)