💡 Problem Formulation: We need to identify the column index of the first occurrence of ‘1’ in each row of a binary matrix, and from these, determine the smallest index (left-most column) where ‘1’ appears. For instance, given the binary matrix [[0,0,1],[1,0,0],[0,1,0]], the desired output is 0 as the first ‘1’ appears in the first column of the second row.
Method 1: Brute Force Approach
This simple brute force method iterates over each row and column, checking every element one-by-one. It is straightforward but not the most efficient as it has a time complexity of O(rows*columns).
Here’s an example:
def find_left_most_column(matrix):
left_most_index = float('inf')
for row in matrix:
for i, val in enumerate(row):
if val == 1:
left_most_index = min(left_most_index, i)
break # Break out of the current row loop
return left_most_index if left_most_index != float('inf') else -1
binary_matrix = [[0,0,1],[1,0,0],[0,1,0]]
print(find_left_most_column(binary_matrix))The output of this code snippet is:
0
This code defines a function find_left_most_column() that takes a binary matrix as input. It initializes the left-most index to infinity, iterates over each row, and updates the left-most index if a ‘1’ is found. If no ‘1’ is found in the entire matrix, it returns -1.
Method 2: Using Python Built-in Functions
Python offers built-in functions like index() to find the first occurrence of an element. This method leverages these functions to improve code readability and potentially, performance.
Here’s an example:
def find_left_most_column_with_builtin(matrix):
left_most_index = float('inf')
for row in matrix:
if 1 in row:
left_most_index = min(left_most_index, row.index(1))
return left_most_index if left_most_index != float('inf') else -1
binary_matrix = [[0,0,1],[1,0,0],[0,1,0]]
print(find_left_most_column_with_builtin(binary_matrix))The output of this code snippet is:
0
Here, the function find_left_most_column_with_builtin() improves upon the brute force approach by using in to check if ‘1’ exists in a row and index() to find its first occurrence. The minimum column index is updated only when ‘1’ is in the row.
Method 3: Using NumPy Library
When working with matrices, the NumPy library offers powerful vectorized operations for better performance. This method uses NumPy to find the left-most ‘1’ in a binary matrix efficiently.
Here’s an example:
import numpy as np
def find_left_most_column_numpy(matrix):
np_matrix = np.array(matrix)
ones_indices = np.where(np_matrix == 1)
return min(ones_indices[1]) if ones_indices[1].size else -1
binary_matrix = [[0,0,1],[1,0,0],[0,1,0]]
print(find_left_most_column_numpy(binary_matrix))The output of this code snippet is:
0
This code converts the list of lists into a NumPy array and then uses the np.where() function to find all indices of ‘1’s. The function then returns the smallest column index. This method is much faster for large matrices due to NumPy’s optimized computations.
Method 4: Transpose and Enumerate
This method involves transposing the matrix so that columns turn into rows, and then using enumeration to find the index of the first row with a ‘1’, which corresponds to the left-most ‘1’ column in the original matrix.
Here’s an example:
def find_left_most_column_transpose(matrix):
transpose = list(zip(*matrix))
for i, row in enumerate(transpose):
if 1 in row:
return i
return -1
binary_matrix = [[0,0,1],[1,0,0],[0,1,0]]
print(find_left_most_column_transpose(binary_matrix))The output of this code snippet is:
0
The function find_left_most_column_transpose() uses the built-in function zip() to transpose the matrix and then iterates over each new row (originally a column) to find the index of the first occurrence of ‘1’.
Bonus One-Liner Method 5: List Comprehension and Min Function
Python’s one-liners can often provide a concise and efficient way to solve problems. This method uses list comprehension combined with the min() function to find the left-most ‘1’.
Here’s an example:
binary_matrix = [[0,0,1],[1,0,0],[0,1,0]] left_most_index = min((row.index(1) for row in binary_matrix if 1 in row), default=-1) print(left_most_index)
The output of this code snippet is:
0
This one-liner initializes the variable left_most_index by finding the smallest index of ‘1’ in any row using list comprehension. The min() function takes a default value to handle cases where ‘1’ is not found.
Summary/Discussion
- Method 1: Brute Force Approach. Easy to understand. Not efficient for large matrices.
- Method 2: Using Python Built-in Functions. Improved readability. Still not the most efficient.
- Method 3: Using NumPy Library. Highly efficient for large data sets. Requires additional dependency.
- Method 4: Transpose and Enumerate. Simple and somewhat efficient. Transposing can be costly for large matrices.
- Method 5: List Comprehension and Min Function. Extremely concise. Readability may suffer for those unfamiliar with the one-liner style.