1582

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:

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.

Companies: Bloomberg, Google

Related Topics:
Array, Matrix

Similar Questions:

• Difference Between Ones and Zeros in Row and Column (Medium)

Hints:

• Keep track of 1s in each row and in each column. Then while iterating over matrix, if the current position is 1 and current row as well as current column contains exactly one occurrence of 1.

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/special-positions-in-a-binary-matrix/
// Author: github.com/lzl124631x
// Time: O(MN * (M + N))
// Space: O(1)
class Solution {
public:
    int numSpecial(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] != 1) continue;
                int row = 0, col = 0;
                for (int x = 0; x < N && row <= 1; ++x) row += A[i][x];
                if (row > 1) continue;
                for (int x = 0; x < M && col <= 1; ++x) col += A[x][j];
                ans += col == 1;
            }
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/special-positions-in-a-binary-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(M + N)
class Solution {
public:
    int numSpecial(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<int> row(M), col(N);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                row[i] += A[i][j];
                col[j] += A[i][j];
            }
        }
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                ans += A[i][j] == 1 && row[i] == 1 && col[j] == 1;
            }
        }
        return ans;
    }
};