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Minimum Number of Flips to Make Binary Grid Palindromic II

Medium

Google

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Topics:

ArraysGreedy Algorithms

You are given an m x n binary matrix grid.

A row or column is considered palindromic if its values read the same forward and backward.

You can flip any number of cells in grid from 0 to 1, or from 1 to 0.

Return the minimum number of cells that need to be flipped to make all rows and columns palindromic, and the total number of 1's in grid divisible by 4.

For example:

grid = [[1,0,0],[0,1,0],[0,0,1]]

In this case, the minimum number of flips is 3.

Here's another example:

grid = [[0,1],[0,1],[0,0]]

In this case, the minimum number of flips is 2.

Explain how you would approach this problem and provide the most efficient solution.

Minimum Number of Flips to Make Binary Grid Palindromic II

Medium

Google

2 views

Topics:

ArraysGreedy Algorithms

You are given an m x n binary matrix grid.

A row or column is considered palindromic if its values read the same forward and backward.

You can flip any number of cells in grid from 0 to 1, or from 1 to 0.

Return the minimum number of cells that need to be flipped to make all rows and columns palindromic, and the total number of 1's in grid divisible by 4.

For example:

grid = [[1,0,0],[0,1,0],[0,0,1]]

In this case, the minimum number of flips is 3.

Here's another example:

grid = [[0,1],[0,1],[0,0]]

In this case, the minimum number of flips is 2.

Explain how you would approach this problem and provide the most efficient solution.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the grid (number of rows and columns), and what is the maximum size they can be?
Are all the values in the grid guaranteed to be either 0 or 1?
What should I return if it's impossible to make the grid palindromic through flips? Should I return -1, null, or throw an exception?
By 'palindromic,' do you mean each row and each column must read the same forwards and backward, or is the entire grid considered as a single sequence?
Can you clarify what a 'flip' operation entails? Does flipping a cell mean toggling its value (0 becomes 1, and 1 becomes 0)?

Brute Force Solution

Approach

The brute force strategy for this problem is to try every possible combination of flips in the grid. For each combination, we check if the resulting grid is palindromic and count the number of flips needed.

Here's how the algorithm would work step-by-step:

Start with the original grid and consider flipping no cells.
Check if the grid is now palindromic. If it is, record the number of flips (which is zero).
Next, consider flipping one cell. Try flipping each cell individually and check if the resulting grid is palindromic after each flip. If it is, record the number of flips (which is one).
Then, consider flipping two cells. Try all possible pairs of cells, flipping both cells in each pair, and check if the resulting grid is palindromic after each flip. If it is, record the number of flips (which is two).
Continue this process, increasing the number of flipped cells each time. For example, after two, we would consider three, four, five, until all the cells are flipped in the grid.
For each combination of flips, determine if the grid is a palindrome.
Once all possible combinations of flips have been explored, select the combination with the fewest flips that makes the grid palindromic.

Code Implementation

def min_flips_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])
    min_flips = float('inf')
    total_cells = rows * cols

    for i in range(2**total_cells):
        temp_grid = [row[:] for row in grid]
        flips_count = 0
        cell_index = 0

        # Iterate through all cells and flip if the corresponding bit is set
        for row_index in range(rows):
            for col_index in range(cols):
                if (i >> cell_index) & 1:
                    temp_grid[row_index][col_index] = 1 - temp_grid[row_index][col_index]
                    flips_count += 1
                cell_index += 1

        # Check if the flipped grid is palindromic
        is_palindromic = True
        for row_index in range(rows):
            for col_index in range(cols // 2):
                if temp_grid[row_index][col_index] != temp_grid[row_index][cols - 1 - col_index]:
                    is_palindromic = False
                    break
            if not is_palindromic:
                break

        if is_palindromic:
            for col_index in range(cols):
                for row_index in range(rows // 2):
                    if temp_grid[row_index][col_index] != temp_grid[rows - 1 - row_index][col_index]:
                        is_palindromic = False
                        break
                if not is_palindromic:
                    break

        # Keep track of minimum flips required
        if is_palindromic:
            min_flips = min(min_flips, flips_count)

    return min_flips

Big(O) Analysis

Time Complexity

O(2^(n*n)) – The brute force approach iterates through all possible combinations of cell flips in the n x n grid. Each cell can either be flipped or not, resulting in 2^(n*n) possible combinations. For each of these combinations, the algorithm checks if the resulting grid is palindromic, which takes O(n*n) time in the worst case. Thus, the overall time complexity is dominated by the enumeration of all possible flip combinations, leading to O(2^(n*n)).

Space Complexity

O(1) – The provided brute force approach explores all possible combinations of flips. While checking each combination, it only modifies the grid in place and doesn't create any auxiliary data structures that scale with the input size. The number of flips tracked and temporary variables used for checking palindromes are constant. Therefore, the space complexity is O(1).

Optimal Solution

Approach

This problem asks for the fewest changes to make a grid symmetric. The clever solution focuses on finding patterns and connections to minimize changes, rather than brute-forcing all possibilities.

Here's how the algorithm would work step-by-step:

First, understand that for a grid to be symmetric, rows and columns need to mirror each other.
Think of the problem as finding the minimum number of mismatches between mirrored positions.
Instead of directly flipping cells, consider comparing mirrored cell pairs to see if they match.
If a mirrored pair doesn't match, it requires a flip. Count these mismatches.
Recognize that you only need to check half of the grid. The other half is determined by the first half due to symmetry.
To avoid re-counting, only count mismatches in the top-left portion of the grid. The mirrored portion of the grid will automatically be correct by flipping the appropriate cells based on the top-left quadrant.
The number of mismatches you find is the minimum number of flips required to make the grid symmetric.

Code Implementation

def min_flips(grid):
    grid_size = len(grid)
    mismatched_pairs_count = 0

    # Iterate over the top-left quadrant of the grid.
    for row_index in range(grid_size // 2):
        for col_index in range((grid_size + 1) // 2):

            # Compare the cell with its mirrored counterpart across rows.
            if grid[row_index][col_index] != grid[grid_size - 1 - row_index][col_index]:
                mismatched_pairs_count += 1

            # Compare the cell with its mirrored counterpart across columns.
            if grid[row_index][col_index] != grid[row_index][grid_size - 1 - col_index]:
                mismatched_pairs_count += 1

            # Compare the cell with its diagonally mirrored counterpart.
            if grid[row_index][col_index] != grid[grid_size - 1 - row_index][grid_size - 1 - col_index]:
                mismatched_pairs_count += 1

    # The number of flips is half the total mismatches.
    # Each flip corrects two mismatches.
    return mismatched_pairs_count // 2

Big(O) Analysis

Time Complexity

O(n²) – The provided solution iterates through the top-left quadrant of the grid to identify mismatched mirrored pairs. The size of the grid is implicitly related to n, where n likely represents the number of rows or columns. The algorithm iterates through approximately n/2 rows and n/2 columns. Therefore, it requires checking roughly (n/2) * (n/2) cell pairs. This simplifies to n² / 4, which in Big O notation is O(n²).

Space Complexity

O(1) – The algorithm's space complexity is dominated by a constant number of integer variables used to track row and column indices during the comparison of mirrored cell pairs. No auxiliary data structures like lists, arrays, or hash maps are created. Thus, the space required remains constant irrespective of the grid size N, where N represents the total number of cells in the grid, simplifying to O(1).

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 flips if the grid is null or empty as it's already a palindrome.
Grid with only one row or column	If only one row or column, the grid is already a palindrome, return 0.
Grid of size 2x2 with opposite corners identical	Handle 2x2 grids efficiently to avoid unnecessary computations when corners match.
Grid where all elements are the same (all 0s or all 1s)	Handle cases where all elements are the same since the number of flips needed will be either 0 or 1.
Grid where no solution (palindrome) is possible	Consider the case where achieving a palindrome is fundamentally impossible due to the arrangement of 0s and 1s, and return -1.
Large grid size exceeding recursion limits	Utilize iterative approaches instead of recursion to avoid stack overflow errors for large grids.
Integer overflow during calculations of row/column indices or flip counts	Use appropriate data types (e.g., long) to prevent integer overflow when calculating indices or flip counts.
Maximum grid size exceeding available memory	Implement space-optimized approaches or algorithms applicable within available memory constraints.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the grid (number of rows and columns), and what is the maximum size they can be?
Are all the values in the grid guaranteed to be either 0 or 1?
What should I return if it's impossible to make the grid palindromic through flips? Should I return -1, null, or throw an exception?
By 'palindromic,' do you mean each row and each column must read the same forwards and backward, or is the entire grid considered as a single sequence?
Can you clarify what a 'flip' operation entails? Does flipping a cell mean toggling its value (0 becomes 1, and 1 becomes 0)?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the original grid and consider flipping no cells.
Check if the grid is now palindromic. If it is, record the number of flips (which is zero).
Next, consider flipping one cell. Try flipping each cell individually and check if the resulting grid is palindromic after each flip. If it is, record the number of flips (which is one).
Then, consider flipping two cells. Try all possible pairs of cells, flipping both cells in each pair, and check if the resulting grid is palindromic after each flip. If it is, record the number of flips (which is two).
Continue this process, increasing the number of flipped cells each time. For example, after two, we would consider three, four, five, until all the cells are flipped in the grid.
For each combination of flips, determine if the grid is a palindrome.
Once all possible combinations of flips have been explored, select the combination with the fewest flips that makes the grid palindromic.

Code Implementation

def min_flips_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])
    min_flips = float('inf')
    total_cells = rows * cols

    for i in range(2**total_cells):
        temp_grid = [row[:] for row in grid]
        flips_count = 0
        cell_index = 0

        # Iterate through all cells and flip if the corresponding bit is set
        for row_index in range(rows):
            for col_index in range(cols):
                if (i >> cell_index) & 1:
                    temp_grid[row_index][col_index] = 1 - temp_grid[row_index][col_index]
                    flips_count += 1
                cell_index += 1

        # Check if the flipped grid is palindromic
        is_palindromic = True
        for row_index in range(rows):
            for col_index in range(cols // 2):
                if temp_grid[row_index][col_index] != temp_grid[row_index][cols - 1 - col_index]:
                    is_palindromic = False
                    break
            if not is_palindromic:
                break

        if is_palindromic:
            for col_index in range(cols):
                for row_index in range(rows // 2):
                    if temp_grid[row_index][col_index] != temp_grid[rows - 1 - row_index][col_index]:
                        is_palindromic = False
                        break
                if not is_palindromic:
                    break

        # Keep track of minimum flips required
        if is_palindromic:
            min_flips = min(min_flips, flips_count)

    return min_flips

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

This problem asks for the fewest changes to make a grid symmetric. The clever solution focuses on finding patterns and connections to minimize changes, rather than brute-forcing all possibilities.

Here's how the algorithm would work step-by-step:

First, understand that for a grid to be symmetric, rows and columns need to mirror each other.
Think of the problem as finding the minimum number of mismatches between mirrored positions.
Instead of directly flipping cells, consider comparing mirrored cell pairs to see if they match.
If a mirrored pair doesn't match, it requires a flip. Count these mismatches.
Recognize that you only need to check half of the grid. The other half is determined by the first half due to symmetry.
To avoid re-counting, only count mismatches in the top-left portion of the grid. The mirrored portion of the grid will automatically be correct by flipping the appropriate cells based on the top-left quadrant.
The number of mismatches you find is the minimum number of flips required to make the grid symmetric.

Code Implementation

def min_flips(grid):
    grid_size = len(grid)
    mismatched_pairs_count = 0

    # Iterate over the top-left quadrant of the grid.
    for row_index in range(grid_size // 2):
        for col_index in range((grid_size + 1) // 2):

            # Compare the cell with its mirrored counterpart across rows.
            if grid[row_index][col_index] != grid[grid_size - 1 - row_index][col_index]:
                mismatched_pairs_count += 1

            # Compare the cell with its mirrored counterpart across columns.
            if grid[row_index][col_index] != grid[row_index][grid_size - 1 - col_index]:
                mismatched_pairs_count += 1

            # Compare the cell with its diagonally mirrored counterpart.
            if grid[row_index][col_index] != grid[grid_size - 1 - row_index][grid_size - 1 - col_index]:
                mismatched_pairs_count += 1

    # The number of flips is half the total mismatches.
    # Each flip corrects two mismatches.
    return mismatched_pairs_count // 2

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 flips if the grid is null or empty as it's already a palindrome.
Grid with only one row or column	If only one row or column, the grid is already a palindrome, return 0.
Grid of size 2x2 with opposite corners identical	Handle 2x2 grids efficiently to avoid unnecessary computations when corners match.
Grid where all elements are the same (all 0s or all 1s)	Handle cases where all elements are the same since the number of flips needed will be either 0 or 1.
Grid where no solution (palindrome) is possible	Consider the case where achieving a palindrome is fundamentally impossible due to the arrangement of 0s and 1s, and return -1.
Large grid size exceeding recursion limits	Utilize iterative approaches instead of recursion to avoid stack overflow errors for large grids.
Integer overflow during calculations of row/column indices or flip counts	Use appropriate data types (e.g., long) to prevent integer overflow when calculating indices or flip counts.
Maximum grid size exceeding available memory	Implement space-optimized approaches or algorithms applicable within available memory constraints.