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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
}
}