Leetcode #1582: Special Positions in a Binary Matrix
In this guide, we solve Leetcode #1582 Special Positions in a Binary Matrix in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Matrix
Intuition
Grid problems are easiest when you define clear row/column boundaries.
A consistent traversal order prevents off-by-one errors.
Approach
Iterate by rows, columns, or layers depending on the requirement.
Keep bounds updated as the traversal progresses.
Steps:
- Define bounds or directions.
- Visit cells in order.
- Update result and move bounds.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.