The Question

Special Positions in a Binary Matrix

Given an M x N binary matrix mat, return the number of special positions in the matrix. A position (i, j) is defined as 'special' if mat[i][j] == 1 and all other elements in the i-th row and j-th column are 0 (i.e., it is the only 1 in its respective row and column). Constraints: - m == mat.length - n == mat[i].length - 1 <= m, n <= 500 - mat[i][j] is either 0 or 1.

Java

Frequency Counting

Two-Pass Matrix Scan

Questions & Insights

Clarifying Questions

What are the constraints on the matrix dimensions $M \times N$?

Assumption:

M

and

N

are between 1 and 500.

O(M \times N)

time complexity is expected.

Can the matrix contain values other than 0 or 1?

Assumption: The problem specifies a "binary matrix," so only 0s and 1s are present.

Is the matrix guaranteed to be non-empty?

Assumption: Yes, the matrix will have at least one element.

Should we optimize for space if the matrix is very large?

Assumption: A standard

O(M+N)

auxiliary space solution is acceptable, but I will consider

O(1)

space as an advanced topic.

Thinking Process

Pre-calculate row and column sums: To determine if a position

(i, j)

is the only 1 in its row and column, we first need to know the total count of 1s in row

i

and column

j

Two-pass approach: In the first pass, iterate through the entire matrix to populate two arrays: rowCount (storing the sum of 1s in each row) and colCount (storing the sum of 1s in each column).

Verification pass: Iterate through the matrix a second time. A position

(i, j)

is "special" if and only if mat[i][j] == 1, rowCount[i] == 1, and colCount[j] == 1.

Complexity alignment: This approach ensures we touch each element exactly twice, resulting in linear time complexity relative to the number of elements in the matrix.

Implementation Breakdown

Problem Set

Functional Requirement: Count all indices

(i, j)

where mat[i][j] == 1 and all other elements in the

i

-th row and

j

-th column are 0.

Input: int[][] mat of size

M \times N

Output: int (total count of special positions).

Constraints:

1 \le M, N \le 500

, mat[i][j] is 0 or 1.

Approach

Algorithm: Frequency Counting (Two-Pass Matrix Scan).

Data Structure: Two 1D Integer Arrays.

Complexity:

Time:

O(M \times N)

to iterate through the matrix.

Space:

O(M + N)

to store row and column sums.

Implementation

Wrap Up

Advanced Topics

Space Optimization ($O(1)$ extra space): If we are allowed to modify the input matrix, we can use the first row and first column to store the counts (similar to "Set Matrix Zeroes"). However, since we are counting 1s and the matrix contains 1s, we'd need a specific encoding or two separate flags to avoid data corruption.

Sparse Matrix Handling: If the matrix is extremely sparse (mostly 0s), we could store the indices of 1s in a list of pairs. We would then only iterate over the list of 1s to verify their row/column uniqueness, which would be

O(K)

where

K

is the number of 1s.

Parallelization: For massive matrices, the first pass (counting) can be easily parallelized using MapReduce principles, where different threads count rows/columns and then aggregate results.