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.

Solution: Sum for each row and column

Brute force:
Time complexity: O(R*C*(R+C))
Space complexity: O(1)

We can pre-compute the sums for each row and each column, ans = sum(mat[r][c] == 1 and rsum[r] == 1 and csum[c] == 1)

Time complexity: O(R*C)
Space complexity: O(R+C)

class Solution {
public:
  int numSpecial(vector<vector<int>>& mat) {
    const int rows = mat.size();
    const int cols = mat[0].size();
    vector<int> rs(rows);
    vector<int> cs(cols);
    for (int r = 0; r < rows; ++r)
      for (int c = 0; c < cols; ++c) {
        rs[r] += mat[r][c];
        cs[c] += mat[r][c];
      }
    int ans = 0;
    for (int r = 0; r < rows; ++r)
      for (int c = 0; c < cols; ++c)
        ans += mat[r][c] && rs[r] == 1 && cs[c] == 1;      
    return ans;
  }
};

The post 花花酱 LeetCode 1582. Special Positions in a Binary Matrix appeared first on Huahua's Tech Road.