The Question

Special Positions in Binary Matrix

Given an m x n binary matrix, a cell (i, j) is defined as 'special' if mat[i][j] is 1 and all other elements in the i-th row and j-th column are 0. Write an efficient algorithm to calculate the total number of special positions in the matrix. Analyze the time and space complexity of your approach.

Java

Pre-Computation

Array

Questions & Insights

Clarifying Questions

What are the constraints on the matrix dimensions $m$ and $n$? (Assumption:

1 \le m, n \le 100

, allowing for an

O(m \cdot n)

or even

O(m \cdot n \cdot (m+n))

solution, though the former is preferred).

Is it guaranteed that the matrix only contains 0s and 1s? (Assumption: Yes, it is a binary matrix).

How should we handle an empty matrix? (Assumption: Based on constraints, the matrix will have at least one element).

Are there any memory constraints? (Assumption: Standard heap limits apply;

O(m+n)

space is perfectly acceptable).

Thinking Process

To identify a "special" position

(i, j)

, we must ensure mat[i][j] == 1 and that no other 1s exist in row

i

and no other 1s exist in column

j

A naive approach would be to iterate through every cell, and if it's a 1, scan its entire row and column. This leads to

O(m \cdot n \cdot (m+n))

complexity.

We can optimize this by pre-calculating the number of 1s in every row and every column. We can use two arrays: rowSums of size

m

and colSums of size

n

Once the sums are pre-calculated, a position

(i, j)

is special if and only if mat[i][j] == 1, rowSums[i] == 1, and colSums[j] == 1. This reduces the complexity to

O(m \cdot n)

Implementation Breakdown

Problem Set

Functional Requirement: Given an

m \times n

binary matrix, return the number of special positions.

Input: int[][] mat

Output: int (count of special positions)

Time Constraint:

O(m \cdot n)

Space Constraint:

O(m + n)

Approach

Algorithm: Frequency Counting / Pre-computation.

Data Structure: Two 1D integer arrays.

Complexity:

Time:

O(m \cdot n)

where

m

is the number of rows and

n

is the number of columns, as we traverse the matrix twice.

Space:

O(m + n)

to store the counts of 1s for each row and column.

Implementation

Wrap Up

Advanced Topics

Readability vs. Performance: The current

O(m \cdot n)

solution is optimal in terms of time complexity for this problem. However, if the matrix is extremely sparse, one could store the indices of 1s in a list of pairs to avoid iterating over many 0s in the second pass.

In-place optimization: If space was strictly

O(1)

and we were allowed to mutate the input, we could potentially encode the counts into the matrix itself (e.g., using bit manipulation or values outside the [0, 1] range), though this often makes the code less readable and harder to maintain.

Parallelization: For massive matrices, the row/column counting phase can be easily parallelized using a Fork-Join framework or MapReduce approach, where different threads count subsets of rows and merge results.