# Leetcode – Special Positions in a Binary Matrix Solution

Matrices are common data structures in computer science and frequently appear in problems related to graphs, dynamic programming, and more. The problem “Special Positions in a Binary Matrix” on Leetcode is a prime example that tests your ability to work with matrices. This problem challenges you to understand and apply array manipulations effectively.

In this extensive article, we will look into the problem’s details, its different approaches, and how to code these solutions in Python. Moreover, we will evaluate each approach based on its time and space complexities.

## Problem Statement

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] is 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

### Constraints:

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

### Example

For mat = [[1,0,0], [0,0,1], [1,0,0]], the answer would be 1.

## Naive Approach: Brute-Force Iteration

### Algorithm

1. Iterate through each cell in the matrix.
2. For each cell that contains a ‘1’, check its entire row and column for other ‘1’s.
3. If no other ‘1’s are found in its row and column, consider it special.

#### Python Implementation:

def numSpecial(mat):
rows, cols = len(mat), len(mat)
count = 0

for i in range(rows):
for j in range(cols):
if mat[i][j] == 1:
if all(mat[i][k] == 0 for k in range(cols) if k != j) and all(mat[k][j] == 0 for k in range(rows) if k != i):
count += 1
return count

### Time Complexity

The time complexity would be O(rows×cols×(rows+cols)).

### Space Complexity

The space complexity is O(1).

## Optimized Approach: Row and Column Count

### Algorithm

1. Pre-calculate the sum of each row and column.
2. Iterate through each cell in the matrix.
3. Check if the pre-calculated sum for its row and column equals 1. If yes, consider it special.

#### Python Implementation:

def numSpecial(mat):
rows, cols = len(mat), len(mat)
rowCount = [sum(mat[i]) for i in range(rows)]
colCount = [sum(mat[i][j] for i in range(rows)) for j in range(cols)]

count = 0
for i in range(rows):
for j in range(cols):
if mat[i][j] == 1 and rowCount[i] == 1 and colCount[j] == 1:
count += 1

return count

### Time Complexity

The time complexity is O(rows×cols).

### Space Complexity

The space complexity is O(rows+cols).

## Conclusion

The “Special Positions in a Binary Matrix” problem serves as a quintessential example of matrix manipulation. It is a well-crafted problem that tests your ability to optimize algorithms for both time and space. Understanding both the naive and optimized approaches provides a comprehensive understanding of how you can improve your problem-solving skills.

While the naive approach is easier to implement, its time complexity makes it less efficient for larger matrices. On the other hand, the optimized approach, although requiring additional space for storing row and column counts, significantly reduces the time complexity, making it the preferred method for larger inputs.