533. Lonely Pixel II (Medium)

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:

1. Row R and column C both contain exactly N black pixels.
2. 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].

Solution:

class Solution {
public:
    int findBlackPixel(vector<vector<char>>& picture, int N) {
        int m = picture.size(), n = picture[0].size();
        vector<int> col(n);

        map<vector<int>, int> m2;

        for (int i = 0; i < m; ++i) {
            vector<int> colpos;
            int cnt = 0;
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B') {
                    colpos.push_back(j);
                    ++cnt;
                    ++col[j];
                }
            }
            if (cnt == N) ++m2[colpos];
        }
        int cnt = 0;
        for (auto& a: m2) {
            if (a.second == N) {
                for (auto& b: a.first) {
                    if (col[b] == N) cnt += N;
                }
            }
        }
        return cnt;
    }
};