Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 13

Description

Special Positions in a Binary Matrix

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:

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.

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 100
• mat[i][j] is either 0 or 1.

Solution

class Solution {
public:
  int numSpecial(vector<vector<int>>& mat) {
    int rows = mat.size();
    int cols = mat.front().size();
    vector<int> rowCount(rows);
    vector<int> colCount(cols);
    for(int i = 0; i < rows; ++i) {
      for(int j = 0; j < cols; ++j) {
        if(mat[i][j]) {
          rowCount[i] += 1;
          colCount[j] += 1;
        }
      }
    }

    int answer = 0;
    for(int i = 0; i < rows; ++i) {
      if(rowCount[i] != 1) continue;
      for(int j = 0; j < cols; ++j) {
        if(!mat[i][j] || colCount[j] != 1) continue;
        answer += 1;
      }
    }

    return answer;
  }
};

// Accepted
// 95/95 cases passed (13 ms)
// Your runtime beats 83.42 % of cpp submissions
// Your memory usage beats 12.89 % of cpp submissions (13.4 MB)