# 533 Lonely Pixel II

### Problem:

Given a picture consisting of black and white pixels, and a positive integer N, find the number of black pixels located at some specific row R and column C that align with all the following rules:

Row R and column C both contain exactly N black pixels. For all rows that have a black pixel at column C, they should be exactly the same as row R The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.

Example:

``````Input:
[['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'W', 'B', 'W', 'B', 'W']]

N = 3
Output: 6
Explanation: All the bold 'B' are the black pixels we need (all 'B's at column 1 and 3).
0    1    2    3    4    5         column index
0    [['W', 'B', 'W', 'B', 'B', 'W'],
1     ['W', 'B', 'W', 'B', 'B', 'W'],
2     ['W', 'B', 'W', 'B', 'B', 'W'],
3     ['W', 'W', 'B', 'W', 'B', 'W']]
row index

Take 'B' at row R = 0 and column C = 1 as an example:
Rule 1, row R = 0 and column C = 1 both have exactly N = 3 black pixels.
Rule 2, the rows have black pixel at column C = 1 are row 0, row 1 and row 2. They are exactly the same as row R = 0.
``````

Note:

1. The range of width and height of the input 2D array is [1,200].

### Solutions:

``````public class Solution {
public int findBlackPixel(char[][] picture, int N) {
char[][] pic = picture;
if (pic == null || pic.length == 0 || pic[0].length == 0) {
return 0;
}
int[] row = new int[pic.length];
int[] col = new int[pic[0].length];
for (int i = 0; i < pic.length; i ++) {
for (int j = 0; j < pic[0].length; j ++) {
if (pic[i][j] == 'B') {
row[i] ++;
col[j] ++;
}
}
}
HashMap<String, List<Integer>> rows = new HashMap<String, List<Integer>>();
for (int i = 0; i < pic.length; i ++) {
if (row[i] != N) {
continue;
}
String arow = new String(pic[i]);
if (!rows.containsKey(arow)) {
}
}
int count = 0;
for (String key:rows.keySet()) {
List<Integer> rowIndex = rows.get(key);
int selected = rowIndex.size();
if (selected != N) {
continue;
}
int i = rowIndex.get(0);
int add = 0;
for (int j = 0; j < pic[0].length; j ++) {
if (pic[i][j] != 'B' || col[j] != N) {
continue;
}
}
count += add * selected;
}
return count;
}
}
``````
``````class Solution {
public int findBlackPixel(char[][] picture, int N) {
char[][] pic = picture;
if (pic == null || pic.length == 0 || pic[0].length == 0) {
return 0;
}
int[] row = new int[pic.length];
int[] col = new int[pic[0].length];
for (int i = 0; i < pic.length; i ++) {
for (int j = 0; j < pic[0].length; j ++) {
if (pic[i][j] == 'B') {
row[i] ++;
col[j] ++;
}
}
}
boolean[] invalidCol = new boolean[pic[0].length];
for (int j = 0; j < invalidCol.length; j ++) {
if (col[j] != N) {
invalidCol[j] = true;
}
}
for (int j = 0; j < pic[0].length; j ++) {
String samerow = null;
System.out.println("checking col " + j);
for (int i = 0; i < pic.length; i ++) {
if (pic[i][j] == 'B') {
if (row[i] != N) {
invalidCol[j] = true;
break;
}
if (samerow == null) {
samerow = new String(pic[i]);
}
else {
if (!samerow.equals(new String(pic[i]))) {
invalidCol[j] = true;
break;
}
}
}
}
}
int count = 0;
for (int j = 0; j < invalidCol.length; j ++) {
if (invalidCol[j] == false) {
count += N;
}
}
return count;
}
}
``````