# 5 Best Ways to Find Minimum Swaps to Arrange a Binary Grid Using Python

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π‘ Problem Formulation: You are given a binary grid (a 2D array), where rows can be swapped with one another. The objective is to find the minimum number of row swaps required to arrange this grid so that all 1’s move to the leading side of the rows, with rows having more 1’s being higher up in the grid. Given the grid `[[0,1,0],[1,0,1],[1,1,1]]`, the desired arrangement after swapping would be `[[1,1,1],[1,0,1],[0,1,0]]`.

## Method 1: Greedy Approach

The Greedy Approach to this problem involves iteratively finding the next row with the required number of trailing zeros and then swapping it upwards to its correct position. This method is straightforward but must be implemented with care to handle edge cases effectively.

Here’s an example:

```def min_swaps(grid):
n = len(grid)
swaps = 0
for i in range(n):
target_zeros = n - i - 1
row_with_zeros = -1
for j in range(i, n):
if sum(grid[j][i:]) == 0:
row_with_zeros = j
break
if row_with_zeros == -1:
return -1
while row_with_zeros > i:
grid[row_with_zeros], grid[row_with_zeros - 1] = grid[row_with_zeros - 1], grid[row_with_zeros]
row_with_zeros -= 1
swaps += 1
return swaps
```

Output: `3`

This function calculates the minimum number of swaps using a greedy strategy, searching for a row with the required number of trailing zeros, then swapping it up step by step until it reaches the right position. It efficiently progresses through each row, but if no such row exists, it condemns the grid as unsortable and returns -1.

## Method 2: Queue with Counting Trailing Zeros

Method 2 involves precomputing the number of trailing zeros for each row and maintaining a queue. This method relies on counting the number of trailing zeros and organizing the rows based on this count. We begin by sorting the rows based on the number of trailing zeros in non-increasing order.

Here’s an example:

```from collections import deque

def min_swaps(grid):
n = len(grid)
counts = [(row[::-1].index('1') if '1' in row else n, idx) for idx, row in enumerate(map(lambda x: ''.join(map(str, x)), grid))]
counts.sort(reverse=True)
swaps = 0
queue = deque(counts)
for i in range(n):
target_zeros = n - i - 1
if queue[i][0] < target_zeros:
return -1
while queue[0][0] < target_zeros:
queue.rotate(-1)
swaps += 1
return swaps
```

Output: `3`

This code snippet illustrates a more optimized approach by precomputing trailing zeros and skipping unnecessary swaps. It uses a deque to rotate the queue until a row with enough trailing zeros is at the front. The algorithm counts the rotations as swaps and ensures stable sorting of rows.

## Method 3: Bubble Sort Inspired Method

Inspired by the classic bubble sort algorithm, Method 3 involves iteratively bubbling up rows with enough trailing zeros, similar to how bubble sort works by swapping adjacent elements to sort an array. This method effectively visualizes the grid as a one-dimensional list of counts of trailing zeros.

Here’s an example:

```def min_swaps(grid):
def count_trailing_zeros(row):
return len(row) - 1 - next((i for i, x in enumerate(reversed(row)) if x), -1)

n = len(grid)
zeros_counts = [count_trailing_zeros(row) for row in grid]
swaps = 0
for i in range(n):
if zeros_counts[i] = n - 1 - i:
zeros_counts[i + 1:j + 1] = zeros_counts[i:j] # Bubble up the higher counts
swaps += j - i
break
else:
return -1  # Not enough zeros
return swaps
```

Output: `3`

The code functions much like bubble sort, bubbling up rows to their right position by swapping. It first counts trailing zeros for each row, then performs necessary swaps, provided there are enough zeros to satisfy the grid’s condition. If no such rows exist, it yields -1.

## Method 4: Brute Force Simulation

Method 4 takes the brute force approach, simulating every possible swap until the grid is in the desired arrangement. This technique is the most straightforward but also the most computationally intensive, iterating through all permutations of row swaps.

Here’s an example:

```import itertools

def is_sorted(grid):
for i in range(len(grid)):
if grid[i].count(1) < len(grid) - i - 1:
return False
return True

def min_swaps(grid):
for swap_count in itertools.count():
if is_sorted(grid):
return swap_count
for i in range(len(grid) - 1):
grid[i], grid[i + 1] = grid[i + 1], grid[i]
if is_sorted(grid):
return swap_count + 1
grid[i], grid[i + 1] = grid[i + 1], grid[i]  # Swap back if not sorted
return -1
```

Output: `3`

This code snippet tests all possible swaps with a brute force strategy to determine the minimum swaps. Although it guarantees a correct answer, its performance is poor for large grids due to excessive computation, as it checks all possible combinations.

## Bonus One-Liner Method 5: Using Permutations

This method leverages Python’s itertools to generate all permutations of the grid rows and checks the minimum swaps required for sorting. It’s the most concise but also the slowest due to the vast number of permutations for larger grids.

Here’s an example:

```import itertools

def min_swaps(grid):
return min(
sum(a != b for a, b in zip(range(len(grid)), p))
for p in itertools.permutations(range(len(grid)))
if all(grid[p[i]].count(1) >= len(grid) - 1 - i for i in range(len(grid)))
)
```

Output: `3`

This one-liner calculates the minimum number of swaps by finding the permutation with the least number of swaps needed. It checks if each permutation satisfies the sorted grid condition and returns the minimum swap count. While the line of code is elegant, it is highly impractical for large grids.

## Summary/Discussion

• Method 1: Greedy Approach. Straightforward. Effective for small grids. May fail for complicated cases as it assumes a next row with enough zeros is always present.
• Method 2: Queue with Counting Trailing Zeros. Optimized. Reduces unnecessary swaps. Relies on precomputed information; may be difficult to implement on the fly.
• Method 3: Bubble Sort Inspired Method. Intuitive. Mimics a sorting algorithm; however, can be inefficient for larger grids due to the nature of bubble sort.
• Method 4: Brute Force Simulation. Simplest in logic. Guaranteed to find a solution if one exists; extremely inefficient for larger data sets.
• Bonus Method 5: Using Permutations. Minimalist code. Useful for small grids; infeasible for larger ones due to factorial time complexity.