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 rowRand columnCthat 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].

分析

zip(*picture)遍历纵行

纵行B总数有N的 row Index push入 数组colb

再行数遍历，有N个B的行，join变成str入rowstr的set, 同时记录共有几个这样的行。

同时满足cnt++ 结果是cnt*N

class Solution:
    def findBlackPixel(self, picture: List[List[str]], N: int) -> int:     
        res = 0

        for col in zip(*picture):
            if col.count('B') == N:
                colb = []
                rowcnt = 0
                rowstr = set()
                for i in range(len(col)):
                    if col[i] == 'B':
                        colb.append(i)
                for i in colb:
                    if picture[i].count('B') == N:
                        rowstr.add(''.join(picture[i]))
                        rowcnt += 1 #B的数目都要是N
                if len(rowstr) == 1 and rowcnt == N:
                    res += 1
        return res*N