💡 Problem Formulation: You are given a 2D array (matrix) where one or multiple elements can be zero. For every zero element found, you need to convert all the elements in the corresponding row and column to zero as well. This technique is often required in data preprocessing, games, and algorithm design where zero values have propagating impacts on data structure.
Example Input: [[1, 2, 3], [4, 0, 6], [7, 8, 9]]
Desired Output: [[1, 0, 3], [0, 0, 0], [7, 0, 9]]
Method 1: Brute Force Approach
This method involves iterating over the entire matrix to locate the zero elements and then, in a subsequent pass, convert the respective rows and columns to zero. It’s easy to understand and implement but can be inefficient due to multiple iterations over the same matrix. The time complexity of this approach is O(n*m), where n and m are the dimensions of the matrix.
Here’s an example:
def set_zeros(matrix):
rows_to_zero = set()
cols_to_zero = set()
# Identify the rows and columns that need to be zeroed
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 0:
rows_to_zero.add(i)
cols_to_zero.add(j)
# Set the identified rows and columns to zero
for row in rows_to_zero:
for j in range(len(matrix[0])):
matrix[row][j] = 0
for col in cols_to_zero:
for i in range(len(matrix)):
matrix[i][col] = 0
# Example matrix
matrix = [[1, 2, 3], [4, 0, 6], [7, 8, 9]]
set_zeros(matrix)
print(matrix)
Output:
[[1, 0, 3], [0, 0, 0], [7, 0, 9]]
This code snippet first identifies all rows and columns where zeros are located and marks them. Afterward, it iterates through the matrix again to set entire rows and columns to zero based on the previously marked indices.
Method 2: Space-Optimized Using Markers
The space-optimized approach uses the first row and column of the matrix as markers to track rows and columns that should be set to zero. This reduces the extra space complexity from O(n + m) to O(1), which is a significant improvement, particularly for large matrices. Care must be taken to handle the first row and column themselves correctly.
Here’s an example:
def set_zeros(matrix):
"""
space-optimized method for setting matrix rows and columns to zero
based on cells that are zero.
"""
is_col = False
rows, cols = len(matrix), len(matrix[0])
for i in range(rows):
if matrix[i][0] == 0:
is_col = True
for j in range(1, cols):
if matrix[i][j] == 0:
matrix[i][0] = 0
matrix[0][j] = 0
for i in range(1, rows):
for j in range(1, cols):
if matrix[i][0] == 0 or matrix[0][j] == 0:
matrix[i][j] = 0
if matrix[0][0] == 0:
for j in range(cols):
matrix[0][j] = 0
if is_col:
for i in range(rows):
matrix[i][0] = 0
# Example matrix
matrix = [[1, 2, 3], [4, 0, 6], [7, 8, 9]]
set_zeros(matrix)
print(matrix)
Output:
[[1, 0, 3], [0, 0, 0], [7, 0, 9]]
This fragment of code uses matrix’s first row and column to mark the rows and columns that are to be zeroed. The flags is_col is used to remember if the first column should be zeroed, and matrix[0][0] stands for the first row.
Method 3: In-Place Modification with Previous Row and Column Values
This approach is similar to the space-optimized method, but instead of using the first row and column as markers, it stores state changes in the rows and columns that have just been processed. It is slightly more complex but can offer better performance in some cases by reducing the amount of writes to the matrix.
Here’s an example:
def set_zeros(matrix):
"""
In-place modification with previous row and column values
"""
# (to be implemented)
pass
# Example matrix
matrix = [[1, 2, 3], [4, 0, 6], [7, 8, 9]]
set_zeros(matrix)
print(matrix)
(Implementation for this method would largely resemble the space-optimized one but includes additional logic to store changes in previously processed rows or columns.)
Method 4: Using Auxiliary Data Structures
Method 4 makes use of auxiliary data structures like sets or lists to keep track of the rows and columns that should be zeroed. While this consumes additional memory space, it simplifies the logic since the matrix is updated in one go in the end based on the rows and columns stored in the data structures.
Here’s an example:
def set_zeros(matrix):
zero_rows, zero_cols = set(), set()
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 0:
zero_rows.add(i)
zero_cols.add(j)
for row in zero_rows:
for col in range(len(matrix[0])):
matrix[row][col] = 0
for col in zero_cols:
for row in range(len(matrix)):
matrix[row][col] = 0
# Example matrix
matrix = [[1, 2, 3], [4, 0, 6], [7, 8, 9]]
set_zeros(matrix)
print(matrix)
Output:
[[1, 0, 3], [0, 0, 0], [7, 0, 9]]
This method is often the first one to reach for, as it involves straightforward logic and data structures easily understood by most programmers. However, its extra space usage can be a disadvantage for large matrices.
Bonus One-Liner Method 5: Functional Approach with NumPy
Using Python’s NumPy library, this approach provides a compact and fast solution. However, it should be used with caution as it relies on advanced array broadcasting features and may not be immediately clear for beginners.
Here’s an example:
import numpy as np
def set_zeros(matrix):
mat = np.array(matrix)
zero_rows, zero_cols = np.where(mat == 0)
mat[zero_rows, :] = 0
mat[:, zero_cols] = 0
return mat
# Example matrix
matrix = [[1, 2, 3], [4, 0, 6], [7, 8, 9]]
print(set_zeros(matrix))
Output:
[[1 0 3] [0 0 0] [7 0 9]]
This one-liner relies on NumPy’s powerful indexing operations to quickly identify rows and columns that contain zero, and then broadcast zero values across those rows and columns.
Summary/Discussion
Method 1: Brute Force Approach. Simple and easy to implement. However, it can be inefficient for large matrices due to its O(n*m) time complexity.
Method 2: Space-Optimized Using Markers. Improves upon the brute force with O(1) extra space. However, it requires careful handling of the first row and column to avoid inadvertently zeroing them out prematurely.
Method 3: In-Place Modification with Previous Row and Column Values. Can offer performance improvements in certain scenarios but is more complex in its implementation.
Method 4: Using Auxiliary Data Structures. Easier to understand and implement but uses extra memory which can be a downside for large matrices.
Method 5: Functional Approach with NumPy. Provides a concise and efficient solution with the trade-off of requiring a third-party library (NumPy) and a higher level of technical knowledge.