# Simple java solution with Rule 2 caveat!

• Record the number of black pixels per row and col. Once you find a row and col that have `N` black pixels, you've found something that obeys rule1 and only need to check if it follows rule2, right?

Not so fast! I discovered (the hard way) that while checking for rule2, all row/col combinations need to again follow rule1. If you fail test case 113, then you've been burned by this (implicit) requirement. Here's a full working solution:

``````    public int findBlackPixel(char[][] picture, int N) {
if (picture == null || picture.length == 0 || picture[0].length == 0 || N < 0) return 0;

int m = picture.length, n = picture[0].length;
int[] rows = new int[m], 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++) {
for (int j = 0; j < n; j++) {
if (picture[i][j] == 'B' && rows[i] == N && cols[j] == N) {
if (obeysRule2(picture, i, j, m, n, rows, cols)) {
res++;
}
}
}
}
return res;
}

private boolean obeysRule2(char[][] picture, int index, int j, int m, int n, int[] rows, int[] cols) {
for (int i = 0; i < m; i++) {
if (i == index) continue;
if (picture[i][j] == 'B') {
if (rows[index] != rows[i]) return false; // violates rule2
for (int col = 0; col < n; col++) {
if (picture[i][col] != picture[index][col]) return false; // violates rule1
}
}
}
return true;
}
``````

• Easy to understand, but terrible efficiency

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