How do you solve Special Positions in a Binary Matrix on LeetCode?

Special Positions in a Binary Matrix (LeetCode #1582) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Counting 1s in rows and columns allows for quick checks.

What is the brute force solution for Special Positions in a Binary Matrix?

The brute force approach for Special Positions in a Binary Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each position in the matrix to see if it is a special position by verifying that it is 1 and that all other elements in its row and column are 0. This is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Special Positions in a Binary Matrix?

Special Positions in a Binary Matrix uses the following concepts: Array, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Special Positions in a Binary Matrix?

Always clarify the problem requirements before jumping into coding. Consider edge cases, such as matrices with no 1s or all 1s. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Special Positions in a Binary Matrix?

Not checking all elements in the row and column for special positions. Overlooking the need to count 1s before checking special positions.

#1582

Special Positions in a Binary Matrix

Easy

ArrayMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking each position multiple times, we can count the number of 1s in each row and column first. This allows us to quickly determine if a position is special in constant time.

⚙️

Algorithm

3 steps

1Step 1: Create two arrays to count the number of 1s in each row and each column.
2Step 2: Iterate through the matrix to populate these counts.
3Step 3: Iterate through the matrix again and check if mat[i][j] == 1 and both rowCounts[i] and colCounts[j] are 1.

solution.py20 lines

1# Full working Python code
2mat = [[1,0,0],[0,0,1],[1,0,0]]
3
4def countSpecialPositions(mat):
5    m, n = len(mat), len(mat[0])
6    rowCounts = [0] * m
7    colCounts = [0] * n
8    for i in range(m):
9        for j in range(n):
10            if mat[i][j] == 1:
11                rowCounts[i] += 1
12                colCounts[j] += 1
13    count = 0
14    for i in range(m):
15        for j in range(n):
16            if mat[i][j] == 1 and rowCounts[i] == 1 and colCounts[j] == 1:
17                count += 1
18    return count
19
20print(countSpecialPositions(mat))

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the matrix a couple of times, and the space complexity is O(n) due to the additional arrays used to store counts.

1Counting 1s in rows and columns allows for quick checks.
2Using additional space can lead to significant time savings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.