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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;
    }
};

  • 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;
    }
}

############

class Solution {
    public int numSpecial(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        int[] r = new int[m];
        int[] c = new int[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                r[i] += mat[i][j];
                c[j] += mat[i][j];
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (mat[i][j] == 1 && r[i] == 1 && c[j] == 1) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

  • // 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;
    }
};

  • class Solution:
    def numSpecial(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        r = [0] * m
        c = [0] * n
        for i, row in enumerate(mat):
            for j, v in enumerate(row):
                r[i] += v
                c[j] += v
        ans = 0
        for i in range(m):
            for j in range(n):
                if mat[i][j] == 1 and r[i] == 1 and c[j] == 1:
                    ans += 1
        return ans



  • func numSpecial(mat [][]int) int {
	m, n := len(mat), len(mat[0])
	r, c := make([]int, m), make([]int, n)
	for i, row := range mat {
		for j, v := range row {
			r[i] += v
			c[j] += v
		}
	}
	ans := 0
	for i, x := range r {
		for j, y := range c {
			if mat[i][j] == 1 && x == 1 && y == 1 {
				ans++
			}
		}
	}
	return ans
}

  • function numSpecial(mat: number[][]): number {
    const m = mat.length;
    const n = mat[0].length;
    const rows = new Array(m).fill(0);
    const cols = new Array(n).fill(0);
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (mat[i][j] === 1) {
                rows[i]++;
                cols[j]++;
            }
        }
    }

    let res = 0;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (mat[i][j] === 1 && rows[i] === 1 && cols[j] === 1) {
                res++;
            }
        }
    }

    return res;
}


  • impl Solution {
    pub fn num_special(mat: Vec<Vec<i32>>) -> i32 {
        let m = mat.len();
        let n = mat[0].len();
        let mut rows = vec![0; m];
        let mut cols = vec![0; n];
        for i in 0..m {
            for j in 0..n {
                rows[i] += mat[i][j];
                cols[j] += mat[i][j];
            }
        }

        let mut res = 0;
        for i in 0..m {
            for j in 0..n {
                if mat[i][j] == 1 && rows[i] == 1 && cols[j] == 1 {
                    res += 1;
                }
            }
        }
        res
    }
}

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