# 531. Lonely Pixel I

## Description

Given an m x n picture consisting of black 'B' and white 'W' pixels, return the number of black lonely pixels.

A black lonely pixel is a character 'B' that located at a specific position where the same row and same column don't have any other black pixels.

Example 1:

Input: picture = [["W","W","B"],["W","B","W"],["B","W","W"]]
Output: 3
Explanation: All the three 'B's are black lonely pixels.


Example 2:

Input: picture = [["B","B","B"],["B","B","W"],["B","B","B"]]
Output: 0


Constraints:

• m == picture.length
• n == picture[i].length
• 1 <= m, n <= 500
• picture[i][j] is 'W' or 'B'.

## Solutions

• class Solution {
public int findLonelyPixel(char[][] picture) {
int m = picture.length, n = picture[0].length;
int[] rows = new int[m];
int[] cols = new int[n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (picture[i][j] == 'B') {
++rows[i];
++cols[j];
}
}
}
int res = 0;
for (int i = 0; i < m; ++i) {
if (rows[i] == 1) {
for (int j = 0; j < n; ++j) {
if (picture[i][j] == 'B' && cols[j] == 1) {
++res;
break;
}
}
}
}
return res;
}
}

• class Solution {
public:
int findLonelyPixel(vector<vector<char>>& picture) {
int m = picture.size(), n = picture[0].size();
vector<int> rows(m);
vector<int> cols(n);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (picture[i][j] == 'B') {
++rows[i];
++cols[j];
}
}
}
int res = 0;
for (int i = 0; i < m; ++i) {
if (rows[i] == 1) {
for (int j = 0; j < n; ++j) {
if (picture[i][j] == 'B' && cols[j] == 1) {
++res;
break;
}
}
}
}
return res;
}
};

• class Solution:
def findLonelyPixel(self, picture: List[List[str]]) -> int:
m, n = len(picture), len(picture[0])
rows, cols = [0] * m, [0] * n
for i in range(m):
for j in range(n):
if picture[i][j] == 'B':
rows[i] += 1
cols[j] += 1
res = 0
for i in range(m):
if rows[i] == 1:
for j in range(n):
if picture[i][j] == 'B' and cols[j] == 1:
res += 1
break
return res


• func findLonelyPixel(picture [][]byte) int {
m, n := len(picture), len(picture[0])
rows := make([]int, m)
cols := make([]int, n)
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if picture[i][j] == 'B' {
rows[i]++
cols[j]++
}
}
}
res := 0
for i := 0; i < m; i++ {
if rows[i] == 1 {
for j := 0; j < n; j++ {
if picture[i][j] == 'B' && cols[j] == 1 {
res++
break
}
}
}
}
return res
}

• function findLonelyPixel(picture: string[][]): number {
const m = picture.length;
const n = picture[0].length;
const rows: number[] = Array(m).fill(0);
const cols: number[] = Array(n).fill(0);
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (picture[i][j] === 'B') {
++rows[i];
++cols[j];
}
}
}
let ans = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (picture[i][j] === 'B' && rows[i] === 1 && cols[j] === 1) {
++ans;
}
}
}
return ans;
}