# 5 Best Ways to Find the Left-Most Column Index With a ‘1’ in a Binary Matrix Using Python

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π‘ 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.