Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.
A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).
 
Example 1:
Input: mat = [[1,0,0],
              [0,0,1],
              [1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:
Input: mat = [[1,0,0],
              [0,1,0],
              [0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions. 

Example 3:
Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:
Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

 
Constraints:

rows == mat.length
cols == mat[i].length
1 <= rows, cols <= 100
mat[i][j] is 0 or 1.

Difficulty: EASY
Starter code: class Solution:
    def numSpecial(self, mat: List[List[int]]) -> int:
        

Solutions:
import itertools

class Solution:

	def numSpecial(self, mat: List[List[int]]) -> int:
		rr = [i for (i, r) in enumerate(mat) if sum(r) == 1]
		cc = [i for (i, c) in enumerate(zip(*mat)) if sum(c) == 1]
		return sum((1 for (i, j) in itertools.product(rr, cc) if mat[i][j] == 1))


class Solution:

	def numSpecial(self, mat: List[List[int]]) -> int:
		mat_t = list(zip(*mat))
		res = 0
		for i in range(len(mat)):
			for j in range(len(mat_t)):
				if mat[i][j] == sum(mat[i]) == sum(mat_t[j]) == 1:
					res += 1
		return res


class Solution:

	def numSpecial(self, mat: List[List[int]]) -> int:
		res = 0
		for i in range(len(mat)):
			for j in range(len(mat[-1])):
				if mat[i][j] == 1 and sum(mat[i]) == 1 and (sum((mat[o][j] for o in range(len(mat)))) == 1):
					res += 1
		return res