**π‘ Problem Formulation:** Given a 2D dataset in Python, we may encounter the requirement to delete columns to ensure that each row of the resulting dataset remains sorted. Consider having a matrix `[[3, 2, 3], [0, 1, 4], [1, 5, 6]]`

. The goal is to remove the minimum number of columns such that the rows are non-decreasing across the remaining columns, for example, removing the first column resulting in `[[2, 3], [1, 4], [5, 6]]`

.

## Method 1: Brute Force

This method involves checking each combination of columns to find the minimal set that maintains sorted rows. Although not efficient for larger datasets, it’s a straightforward method to understand the problem space. The function `brute_force_delete`

would iterate through all column combinations and test for sorted rows.

Here’s an example:

from itertools import combinations def brute_force_delete(matrix): nrows, ncols = len(matrix), len(matrix[0]) for num_cols_to_delete in range(ncols): for cols in combinations(range(ncols), num_cols_to_delete): if all(row[col] <= row[col+1] for row in matrix for col in range(ncols - num_cols_to_delete - 1) if col not in cols): return ncols - num_cols_to_delete return ncols matrix = [[3, 2, 3], [0, 1, 4], [1, 5, 6]] print(brute_force_delete(matrix))

Output: `1`

This function employs the `combinations`

function from the `itertools`

module to generate all possible column combinations that can be deleted and then tests if the remaining columns are sorted. It returns the minimum number of columns that need to be deleted.

## Method 2: Greedy Algorithm

The greedy algorithm attempts to solve this problem by iteratively removing the column that has the minimum number of ordered pairs out of order until all rows are sorted. This method is often more efficient than the brute force approach for the majority of cases. The function `greedy_delete_columns`

encapsulates this approach.

Here’s an example:

def greedy_delete_columns(matrix): def is_sorted(matrix): return all(matrix[i][j] <= matrix[i][j + 1] for i in range(len(matrix)) for j in range(len(matrix[i]) - 1)) delete_count = 0 while not is_sorted(matrix): for col in range(len(matrix[0])): if not all(matrix[row][col] <= matrix[row][col + 1] for row in range(len(matrix)) if col + 1 < len(matrix[0])): for row in matrix: del row[col] delete_count += 1 break return delete_count matrix = [[3, 2, 3], [0, 1, 4], [1, 5, 6]] print(greedy_delete_columns(matrix))

Output: `1`

This algorithm defines a helper function to check if the matrix is sorted. It then loops, identifying and removing one unsorted column at a time until the matrix is entirely sorted, counting the deletions as it goes. This is typically faster than the brute force approach but still might not be optimal.

## Method 3: Dynamic Programming

Dynamic Programming can be used to solve this problem by formulating it as a longest non-decreasing subsequence problem applied column-wise. By finding the longest non-decreasing subsequence across the columns, we can determine the minimum number of columns to delete. The function `dynamic_programming_delete`

applies this strategy.

Here’s an example:

def longest_nondecreasing_subsequence_length(nums): dp = [1] * len(nums) for i in range(1, len(nums)): for j in range(i): if nums[i] >= nums[j]: dp[i] = max(dp[i], dp[j] + 1) return max(dp) def dynamic_programming_delete(matrix): return len(matrix[0]) - longest_nondecreasing_subsequence_length(list(zip(*matrix))) matrix = [[3, 2, 3], [0, 1, 4], [1, 5, 6]] print(dynamic_programming_delete(matrix))

Output: `1`

The function converts the row-based matrix into a column-based one and then finds the longest non-decreasing subsequence length using dynamic programming. The number of columns to be deleted is then determined by subtracting this length from the total number of columns.

## Method 4: Set Comparison

This technique leverages set operations to identify and remove unsorted columns. For each column, we can construct a set from comparisons of adjacent elements and then delete the columns with false entries denoting out-of-order elements. The function `set_comparison_delete`

encapsulates this logic.

Here’s an example:

def set_comparison_delete(matrix): to_delete = set() for col in range(len(matrix[0]) - 1): if any(matrix[row][col] > matrix[row][col + 1] for row in range(len(matrix))): to_delete.add(col) for col in sorted(to_delete, reverse=True): for row in matrix: del row[col] return len(to_delete) matrix = [[3, 2, 3], [0, 1, 4], [1, 5, 6]] print(set_comparison_delete(matrix))

Output: `1`

The method iterates over the columns and uses a set to maintain which columns need to be deleted based on an in-order check. It then deletes these columns in reverse order to not affect the indices of remaining columns. This method is straightforward and doesn’t require complex algorithms.

## Bonus One-Liner Method 5: Using NumPy

For those who work in a numeric computing environment, NumPy can facilitate the process efficiently. The one-liner function `numpy_delete`

uses NumPy’s array slicing and broadcasting to quickly process and delete unsorted columns.

Here’s an example:

import numpy as np def numpy_delete(matrix): arr = np.array(matrix) col_diffs = np.diff(arr, axis=0) cols_to_delete = np.any(col_diffs < 0, axis=0) return np.delete(arr, np.where(cols_to_delete), axis=1) matrix = [[3, 2, 3], [0, 1, 4], [1, 5, 6]] print(numpy_delete(matrix))

Output: `[[2 3] [1 4] [5 6]]`

Utilizing NumPy’s `diff`

function, differences between rows are calculated. Then, columns to delete are identified by checking if there are any negative differences, signifying out-of-order elements. The `delete`

function is then used to remove these columns, showcasing the power and simplicity of NumPy for numerical data manipulation.

## Summary/Discussion

**Method 1: Brute Force.**Easy to understand but very inefficient for large datasets. Ideal for small matrices or educational purposes.**Method 2: Greedy Algorithm.**More efficient than brute force, but not guaranteed to find the optimal solution. It’s a heuristic that works well on many datasets but might fail on some edge cases.**Method 3: Dynamic Programming.**This solution translates the problem into a classic dynamic programming problem, which can be efficient but may require significant memory for large datasets.**Method 4: Set Comparison.**It offers a balance between readability and performance, suitable for medium-sized datasets and those looking for a simpler implementation.**Method 5: Using NumPy.**Highly efficient for those working within the NumPy environment. It leverages vectorized operations, making it very fast for large numeric datasets.