💡 Problem Formulation: In a binary matrix, a special position is defined as an element that is the only ‘1’ in its row and column. Given a binary matrix, the task is to count the total number of such special positions. For instance, consider the binary matrix
[[1, 0, 0], [0, 0, 1], [0, 1, 0]]Here, there are two special positions: matrix[0][0] and matrix[1][2].
Method 1: Brute Force Check
This approach involves iterating through each element of the matrix, checking if it is equal to ‘1’, and then confirming if it’s the only ‘1’ in its row and column. The function find_special_positions(matrix) performs this check for the entire matrix and returns the count of special positions.
Here’s an example:
def find_special_positions(matrix):
num_of_special_positions = 0
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 1:
if matrix[i].count(1) == 1 and all(matrix[k][j] == 0 for k in range(len(matrix)) if k != i):
num_of_special_positions += 1
return num_of_special_positions
# Example matrix
binary_matrix = [[1, 0, 0], [0, 0, 1], [0, 1, 0]]
print(find_special_positions(binary_matrix))
Output:
2
In this code snippet, we define a function that iterates over each element in the matrix. It checks if an element is ‘1’ and whether it is the only ‘1’ in its row and column. If these conditions are met, it is counted as a special position. Finally, the function returns the total count.
Method 2: Use NumPy for Efficient Operations
For matrices, the NumPy library can simplify operations and make them more efficient. The numpy.where() function along with row and column sum computations are used to identify and count special positions.
Here’s an example:
import numpy as np
def find_special_positions_numpy(matrix):
matrix = np.array(matrix)
rows, cols = np.sum(matrix, axis=1), np.sum(matrix, axis=0)
return sum(matrix[i, j] and rows[i] == cols[j] == 1 for i, j in zip(*np.where(matrix == 1)))
# Example matrix
binary_matrix = [[1, 0, 0], [0, 0, 1], [0, 1, 0]]
print(find_special_positions_numpy(binary_matrix))
Output:
2
This method takes the binary matrix as a NumPy array. We calculate the sums of each row and each column and store them. Then, we iterate over the positions where the matrix value is ‘1’ and check if the sum of the corresponding row and column is equal to 1. The function returns the count of such special positions.
Method 3: Row and Column Dictionaries
By pre-computing the count of ‘1’s in each row and column using dictionaries, we can optimize the lookup time for subsequent checks on each ‘1’. The function special_positions_with_dicts(matrix) implements this method.
Here’s an example:
def special_positions_with_dicts(matrix):
num_of_special_positions = 0
rows = [0]*len(matrix)
cols = [0]*len(matrix[0])
for i in range(len(matrix)):
for j in range(len(matrix[0])):
rows[i] += matrix[i][j]
cols[j] += matrix[i][j]
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 1 and rows[i] == cols[j] == 1:
num_of_special_positions += 1
return num_of_special_positions
# Example matrix
binary_matrix = [[1, 0, 0], [0, 0, 1], [0, 1, 0]]
print(special_positions_with_dicts(binary_matrix))
Output:
2
This function creates two lists to keep the count of ‘1’s in each row and column. It first computes these counts and then iterates over the matrix to check for special positions where the element is ‘1’, and both the corresponding row and column counts are equal to 1.
Method 4: Improved Brute Force with Early Stopping
An improved version of the brute force method stops the row or column search as soon as more than one ‘1’ is found. The find_special_positions_improved(matrix) function implements this optimization.
Here’s an example:
def find_special_positions_improved(matrix):
def is_only_one(m_list):
return m_list.count(1) == 1
special_count = 0
for r_idx, row in enumerate(matrix):
for c_idx, value in enumerate(row):
if value == 1:
column = [matrix[i][c_idx] for i in range(len(matrix))]
if is_only_one(row) and is_only_one(column):
special_count += 1
return special_count
# Example matrix
binary_matrix = [[1, 0, 0], [0, 0, 1], [0, 1, 0]]
print(find_special_positions_improved(binary_matrix))
Output:
2
The code defines an inner function is_only_one() to verify if a list contains only a single ‘1’. We iterate through the matrix, and for each ‘1’, we create a list representing its column. Then, we check if both the current row and this column contain exactly one ‘1’. If so, it is a special position.
Bonus One-Liner Method 5: Python List Comprehension with All()
Using the power of Python list comprehension and the all() function, this one-liner solution performs an efficient single-pass check to find the special positions in a binary matrix.
Here’s an example:
def find_special_positions_oneliner(matrix):
return sum(all(x == 0 for k, x in enumerate(row) if k != j) and all(matrix[i][j] == 0 for i in range(len(matrix)) if i != row_idx) for row_idx, row in enumerate(matrix) for j in range(len(row)) if row[j] == 1)
# Example matrix
binary_matrix = [[1, 0, 0], [0, 0, 1], [0, 1, 0]]
print(find_special_positions_oneliner(binary_matrix))
Output:
2
The one-liner function iterates over every element in the matrix. For each ‘1’ found, it uses list comprehensions to check if it is the only ‘1’ in its row and column. It sums the boolean results of these checks to get the total number of special positions.
Summary/Discussion
- Method 1: Brute Force Check. Simple and direct approach. However, it can be inefficient for large matrices due to nested loops and repeated counts.
- Method 2: Use NumPy for Efficient Operations. Highly efficient for large datasets. Requires NumPy, may not be the best for those who prefer standard library solutions.
- Method 3: Row and Column Dictionaries. Improves lookup times using precomputation. Still, it suffers from having to perform two full matrix scans.
- Method 4: Improved Brute Force with Early Stopping. Offers optimization over basic brute force by reducing unnecessary computations. Still less efficient than NumPy-based solutions.
- Bonus Method 5: Python List Comprehension with All(). Elegant and Pythonic, though it may sacrifice some readability for conciseness. Efficiency similar to that of improved brute force.