# 531 Lonely Pixel I

Given a picture consisting of black and white pixels, find the number ofblacklonely pixels.
The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.
A black lonely pixel is character 'B' that located at a specific position where the same row and same column don't have any other black pixels.
Example:
Input:
[['W', 'W', 'B'],
['W', 'B', 'W'],
['B', 'W', 'W']]
Output:
3
Explanation:
All the three 'B's are black lonely pixels.
Note:
1. 1.
The range of width and height of the input 2D array is [1,500].
The Idea: In the first run through the picture, collect the number of black pixels on each row and column. Then on the second run, verify every black pixel to have a count of 1 in both it's respected row and column.
Complexity: O(n+m) time and space with `n = row count` and `m = col count`
int findLonelyPixel(vector<vector<char>>& picture) {
if (picture.empty()) return 0;
const size_t rows = picture.size();
const size_t cols = picture.at(0).size();
vector<int> rowsWithB(rows);
vector<int> colsWithB(cols);
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (picture[i][j] == 'B') {
rowsWithB[i]++; colsWithB[j]++;
}
}
}
int count = 0;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (picture[i][j] == 'B' && rowsWithB[i] == 1 && colsWithB[j] == 1) {
count++;
}
}
}
return count;
}