533. Lonely Pixel II

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

Solution：

思路：
给了一个整数N，说对于均含有N个黑像素的某行某列，如果该列中所有的黑像素所在的行都相同的话，该列的所有黑像素均为孤独的像素，让我们统计所有的这样的孤独的像素的个数。

那么跟之前那题类似，我们还是要统计每一行每一列的黑像素的个数，而且由于条件二中要比较各行之间是否相等，如果一个字符一个字符的比较写起来比较麻烦，我们可以用个trick，把每行的字符连起来，形成一个字符串，然后直接比较两个字符串是否相等会简单很多(用hashmap(string, count))。然后我们遍历每一行和每一列，如果某行和某列的黑像素刚好均为N，我们遍历该列的所有黑像素，如果其所在行均相等，则说明该列的所有黑像素均为孤独的像素，将个数加入结果res中，然后将该行的黑像素统计个数清零，以免重复运算，这样我们就可以求出所有的孤独的像素了，

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

// 孤独的点要满足:
// (1) 该行有N个黑点
// (2) 该列有N个黑点
// (3) 该列中所有的黑像素所在的行都相同
public class Solution {
    public int findBlackPixel(char[][] picture, int N) {
        if(picture == null || picture.length == 0 || picture[0].length == 0) return 0;
        int m = picture.length;
        int n = picture[0].length;
        
        Map map = new HashMap<>();
        int[] colCount = new int[n];
        
        // (预备3)行编码存入map，(预备2)统计列的B，并check(1)该行B的个数是否为N
        for (int i = 0; i < m; i++) {
            String key = scanRow(picture, i, N, colCount);
            if (key.length() != 0) {
                map.put(key, map.getOrDefault(key, 0) + 1);
            }
        }
        
        // check(3) 及 check(2)该列B的个数是否为N
        int result = 0;
        for (String key : map.keySet()) {
            if (map.get(key) == N) {
                for (int j = 0; j < n; j++) {
                    if (key.charAt(j) == 'B' && colCount[j] == N) {
                        result += N;
                    }
                }
            }
        }
        
        return result;
    }
    
    private String scanRow(char[][] picture, int row, int target, int[] colCount) {
        int n = picture[0].length;
        int rowCount = 0;
        StringBuilder sb = new StringBuilder();
        
        for (int j = 0; j < n; j++) {
            if (picture[row][j] == 'B') {
                rowCount++;
                colCount[j]++;
            }
            sb.append(picture[row][j]);
        }
        
        if (rowCount == target) return sb.toString();
        return "";
    }
}