Formatted question description: https://leetcode.ca/all/1582.html

1582. Special Positions in a Binary Matrix (Easy)

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.

Related Topics:
Array

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/special-positions-in-a-binary-matrix/

// 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/

// 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;
    }
};

Java

  • class Solution {
        public int numSpecial(int[][] mat) {
            int count = 0;
            int rows = mat.length, columns = mat[0].length;
            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < columns; j++) {
                    if (mat[i][j] == 1) {
                        if (isSpecial(mat, i, j))
                            count++;
                    }
                }
            }
            return count;
        }
    
        public boolean isSpecial(int[][] mat, int row, int column) {
            int rows = mat.length, columns = mat[0].length;
            for (int i = 0; i < rows; i++) {
                if (i == row)
                    continue;
                if (mat[i][column] == 1)
                    return false;
            }
            for (int j = 0; j < columns; j++) {
                if (j == column)
                    continue;
                if (mat[row][j] == 1)
                    return false;
            }
            return true;
        }
    }
    
  • // OJ: https://leetcode.com/problems/special-positions-in-a-binary-matrix/
    // 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;
        }
    };
    
  • print("Todo!")
    

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