README.md

1582. Special Positions in a Binary Matrix

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, 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:

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.

Example 3:

Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:

Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

Constraints:

• rows == mat.length
• cols == mat[i].length
• 1 <= rows, cols <= 100
• mat[i][j] is 0 or 1.

Solutions (Rust)

1. Count

impl Solution {
    pub fn num_special(mat: Vec<Vec<i32>>) -> i32 {
        let rows = mat.len();
        let cols = mat[0].len();
        let mut row1 = vec![0; rows];
        let mut col1 = vec![0; cols];
        let mut ret = 0;

        for r in 0..rows {
            for c in 0..cols {
                if mat[r][c] == 1 {
                    row1[r] += 1;
                    col1[c] += 1;
                }
            }
        }

        for r in 0..rows {
            for c in 0..cols {
                if mat[r][c] == 1 && row1[r] + col1[c] == 2 {
                    ret += 1;
                }
            }
        }

        ret
    }
}