#1582

Easy

LeetCode Problem Workspace

Special Positions in a Binary Matrix

Identify all special positions in a binary matrix where a 1 has no other 1s in its row or column, ensuring optimal array traversal.

array matrix

Problem Statement

Given an m x n binary matrix mat, return the total number of special positions. A position (i, j) is special if mat[i][j] equals 1 and all other elements in row i and column j are 0. Rows and columns are zero-indexed.

For example, in mat = [[1,0,0],[0,0,1],[1,0,0]], the position (1,2) is special because it is 1 and no other elements in its row or column are 1. Your task is to count all such positions efficiently, respecting the array plus matrix pattern and avoiding unnecessary iteration.

Examples

Example 1

Input: mat = [[1,0,0],[0,0,1],[1,0,0]]

Output: 1

(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

(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.

Solution Approach

Track Row and Column Counts

First, traverse the matrix once to count the number of 1s in each row and each column. Use two arrays rowCount and colCount to store these totals, aligning with the array plus matrix pattern for clear bookkeeping.

Iterate to Identify Special Positions

Next, iterate through every cell in the matrix. If a cell is 1 and rowCount[i] and colCount[j] both equal 1, then the position is special. Increment a running total to count all valid special positions.

Optimize Space and Performance

This approach avoids extra nested loops by using precomputed counts. It achieves O(m * n) time and O(m + n) space complexity, maintaining efficiency for large matrices while preventing common pitfalls like repeated scanning of rows and columns.

Complexity Analysis

Metric	Value
Time	O(m \cdot n)
Space	O(m + n)

Time complexity is O(m * n) because each element is visited twice: once for counting and once for validation. Space complexity is O(m + n) due to the rowCount and colCount arrays storing the number of 1s per row and column.

What Interviewers Usually Probe

Looking for awareness of how to track multiple constraints simultaneously across rows and columns.
Checking if candidate recognizes the array plus matrix pattern to avoid repeated scanning.
Observing if candidate correctly identifies the special position condition without unnecessary nested loops.

Common Pitfalls or Variants

Common pitfalls

Scanning rows and columns repeatedly instead of precomputing counts, leading to O(m * n * max(m,n)) complexity.
Forgetting that a position is only special if both its row and column have exactly one 1.
Modifying the matrix in-place which can invalidate other checks and result in incorrect counts.

Follow-up variants

Counting special positions in a rectangular binary matrix with unequal row and column lengths.
Extending to k-special positions where exactly k 1s are allowed in a row or column.
Identifying special positions in a sparse matrix stored in coordinate list format to reduce memory overhead.

FAQ

What defines a special position in a binary matrix?

A special position is any cell mat[i][j] that equals 1 and has no other 1s in its entire row i and column j.

How do row and column counts help in solving this problem?

By precomputing the number of 1s in each row and column, you can efficiently check the special position condition without scanning rows and columns repeatedly.

Can this approach handle large matrices efficiently?

Yes, the method runs in O(m * n) time with O(m + n) extra space, making it suitable for matrices up to 100x100 within constraints.

Are there variations of the special positions problem?

Yes, you can extend it to k-special positions, rectangular matrices, or sparse storage formats, all relying on the same row and column counting logic.

Why is this an 'Array plus Matrix' pattern problem?

Because the solution combines matrix traversal with array bookkeeping for rows and columns, exemplifying the array plus matrix pattern clearly.

terminal

Solution

Solution 1: Counting

We can use two arrays, $\textit{rows}$ and $\textit{cols}$, to record the number of $1$s in each row and each column, respectively.

1

2

3

4

5

6

7

8

9

10

11

12

13

class Solution:
    def numSpecial(self, mat: List[List[int]]) -> int:
        rows = [0] * len(mat)
        cols = [0] * len(mat[0])
        for i, row in enumerate(mat):
            for j, x in enumerate(row):
                rows[i] += x
                cols[j] += x
        ans = 0
        for i, row in enumerate(mat):
            for j, x in enumerate(row):
                ans += x == 1 and rows[i] == 1 and cols[j] == 1
        return ans

Continue Practicing

#1591 strange printer ii #1572 matrix diagonal sum #1594 maximum non negative product in a matrix