1582. Special Positions in a Binary Matrix

#### QUESTION:

Given an m x n binary matrix mat, 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.
``````

Constraints:

``````m == mat.length
n == mat[i].length
1 <= m, n <= 100
mat[i][j] is either 0 or 1.
``````

#### EXPLANATION:

for循环遍历每一个数, 如果是1, 就拿出来看看改行该列是不是只有这一个. 是的话就结果+1

#### SOLUTION:

``````class Solution {
func numSpecial(_ mat: [[Int]]) -> Int {
var result:Int = 0
for indexI in mat.indices {
for indexJ in mat[indexI].indices {
if mat[indexI][indexJ] == 1 {
if numSpecialHelper(mat, indexI: indexI, IndexJ: indexJ) {
result += 1
}
}
}
}
return result
}

func numSpecialHelper(_ mat: [[Int]], indexI:Int , IndexJ:Int) -> Bool {
var result:Bool = true
var sum:Int = 0
for num in mat[indexI] {
sum += num
}
if sum != 1 {
return false
}
sum = 0
for indexC in 0...mat.count - 1 {
sum += mat[indexC][IndexJ]
}
if sum != 1 {
return false
}
return result
}
}
``````
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