Is Special Positions in a Binary Matrix easy or hard?

LeetCode classifies this problem as Easy because it mainly requires matrix traversal and simple counting. The key idea is recognizing that preprocessing rows and columns avoids repeated checks and improves performance.

Special Positions in a Binary Matrix Python/Java solution

Most implementations follow the same logic: compute row and column counts, then verify each cell. The approach translates directly into Python, Java, C++, and other languages using simple loops and integer arrays.

How to solve Special Positions in a Binary Matrix in O(n)?

For an m x n matrix, the optimal solution is O(m*n). First compute row and column counts in one pass, then check each cell in a second pass. A position is special if it contains 1 and both its row and column counts equal 1.

What is the best approach for Special Positions in a Binary Matrix?

The best approach is row and column preprocessing. Count the number of 1s in each row and each column using two arrays, then scan the matrix again and check if mat[i][j] == 1 with rowCount[i] == 1 and colCount[j] == 1. This reduces the complexity to O(m*n) time with O(m+n) extra space.

Is Special Positions in a Binary Matrix asked at Google/Amazon/Meta?

This problem represents a common matrix counting pattern used in interviews at companies like Amazon and Google. Variations often appear where candidates must preprocess rows or columns to avoid repeated scans.

What data structure is used in Special Positions in a Binary Matrix?

The solution primarily uses arrays for row and column counting along with standard matrix traversal. Two auxiliary arrays store the number of 1s in each row and column to enable constant-time checks.

What is the time complexity of Special Positions in a Binary Matrix?

The optimal solution runs in O(m*n) time because each cell in the matrix is visited a constant number of times. A brute-force approach that scans the entire row and column for every 1 takes O(m*n*(m+n)) time.

Description Hints 1 Solution Video 16:10

1582. Special Positions in a Binary Matrix

Easy72.6% acceptance3 companies

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.

Asked at 3 companies

Etsy

Google Amazon

Topics

Array Matrix

Hints open one at a time

Hint 1

Keep track of 1s in each row and in each column. Then while iterating over matrix, if the current position is 1 and current row as well as current column contains exactly one occurrence of 1.

16:10

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution