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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.
class Solution:
def numSpecial(self, mat: List[List[int]]) -> int:
m = len(mat)
n = len(mat[0])
res = 0
for i in range(m):
for j in range(n):
if mat[i][j] == 1:
count1 = 0
count2 = 0
for k in range(n):
if mat[i][k] == 0:
count1 += 1
for k in range(m):
if mat[k][j] == 0:
count2 += 1
if count1 == n - 1 and count2 == m - 1:
res += 1
return res
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