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Check if There is a Path With Equal Number of 0's And 1's

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Google

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Topics:

Dynamic Programming

You are given a 0-indexed m x n binary matrix grid.

You can move from a cell (row, col) to any of the cells (row + 1, col) or (row, col + 1).

Return true if there is a path from (0, 0) to (m - 1, n - 1) such that the number of zeros and ones are equal. Otherwise, return false.

Example 1:

Input: grid = [[0,1,0,0],[0,1,0,0],[1,0,1,0]]
Output: true
Explanation: One possible path is shown in the diagram. 
The number of zeros (4) and ones (4) are equal.

Example 2:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,0,0,1]]
Output: false
Explanation: There is no path from (0, 0) to (m - 1, n - 1) such that the number of zeros and ones are equal.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 30
grid[i][j] is either 0 or 1.

Try coding it in LeetCode

Check if There is a Path With Equal Number of 0's And 1's

Medium

Google

4 views

Topics:

Dynamic Programming

You are given a 0-indexed m x n binary matrix grid.

You can move from a cell (row, col) to any of the cells (row + 1, col) or (row, col + 1).

Return true if there is a path from (0, 0) to (m - 1, n - 1) such that the number of zeros and ones are equal. Otherwise, return false.

Example 1:

Input: grid = [[0,1,0,0],[0,1,0,0],[1,0,1,0]]
Output: true
Explanation: One possible path is shown in the diagram. 
The number of zeros (4) and ones (4) are equal.

Example 2:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,0,0,1]]
Output: false
Explanation: There is no path from (0, 0) to (m - 1, n - 1) such that the number of zeros and ones are equal.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 30
grid[i][j] is either 0 or 1.

Try coding it in LeetCode

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, and what is the maximum possible size for the grid's dimensions?
What values can the grid cells contain besides 0 and 1? Are negative numbers or null values possible?
Can I only move right and down in the grid, or are other movement directions allowed?
If no path exists with an equal number of 0's and 1's, what should the function return? Should I return null, false, or throw an exception?
Is there any limit on the number of steps/cells I can take in the path, or is the only restriction to reach the destination?

Brute Force Solution

Approach

The brute force method exhaustively explores every possible path through the grid. For each path, it counts the number of zeros and ones encountered. It then checks if that count satisfies the condition of having an equal number of zeros and ones.

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

Start at the beginning of the grid.
Explore every possible path from the starting point to the end of the grid.
For each path, keep track of how many zeros and how many ones you see along the way.
After you reach the end of the grid for a particular path, check if the number of zeros you counted is equal to the number of ones you counted.
If they are equal, you've found a valid path.
Continue exploring all other possible paths, repeating the counting and checking steps.
If any path has an equal number of zeros and ones, you can say that a valid path exists.

Code Implementation

def path_with_equal_zeros_and_ones_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    def explore_path(row_index, column_index, current_path_zeros, current_path_ones):
        #If we reach the end, check if 0s equal 1s
        if row_index == number_of_rows - 1 and column_index == number_of_columns - 1:
            if matrix[row_index][column_index] == 0:
                current_path_zeros += 1
            else:
                current_path_ones += 1
            return current_path_zeros == current_path_ones

        if matrix[row_index][column_index] == 0:
            current_path_zeros += 1
        else:
            current_path_ones += 1

        # Explore right and down if possible
        path_exists = False
        if row_index + 1 < number_of_rows:
            path_exists = path_exists or explore_path(row_index + 1, column_index, current_path_zeros, current_path_ones)

        if column_index + 1 < number_of_columns:
            path_exists = path_exists or explore_path(row_index, column_index + 1, current_path_zeros, current_path_ones)

        return path_exists

    #Start at the top left (0,0)
    return explore_path(0, 0, 0, 0)

Big(O) Analysis

Time Complexity

O(2^(m+n)) – The brute force approach explores every possible path in the m x n grid. From the starting point (0, 0), we can only move right or down. In the worst case, to reach the destination (m-1, n-1), we must make m-1 downward moves and n-1 rightward moves, resulting in a total of m+n-2 moves. For each move, we have two choices (down or right), leading to approximately 2^(m+n-2) possible paths. Therefore, the time complexity is O(2^(m+n)).

Space Complexity

O(2^(m+n)) – The brute force approach explores every possible path from the top-left to the bottom-right of the m x n grid. In the worst case, the algorithm might explore nearly every possible path. The recursion depth can be up to m+n, but more significantly, the number of possible paths grows exponentially, and although not stored explicitly, the computation to get to each path requires recalculation. Therefore, considering the implicit memory use due to the exponential number of paths explored, the space complexity is O(2^(m+n)), where m and n are the dimensions of the grid.

Optimal Solution

Approach

The core idea is to track the difference between the number of 1s and 0s encountered as we move through the grid. We can use this difference to determine if a path with an equal number of 0s and 1s exists.

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

Start at the top-left corner of the grid and initialize a counter to zero. This counter represents the difference between 1s and 0s.
As you move from one cell to the next, either down or to the right, update the counter. If the cell contains a '1', increase the counter; if it contains a '0', decrease it.
Keep track of all the counter values you've seen along the way. If you ever revisit a counter value, it means you can reach that point with multiple different paths having the same difference between 1s and 0s.
When you reach the bottom-right corner, check if the counter is zero. A zero counter means you've found a path where the number of 1s and 0s are equal.
To make it efficient, we can stop exploring a path if it's impossible to reach zero at the end. We can do this by checking if remaining steps can compensate for current difference between the 1s and 0s.

Code Implementation

def has_valid_path(grid):
    rows = len(grid)
    columns = len(grid[0])

    def depth_first_search(row_index, column_index, difference):
        # Base case: Reached the destination.
        if row_index == rows - 1 and column_index == columns - 1:
            return difference + (1 if grid[row_index][column_index] == 1 else -1) == 0

        # Pruning: Check if the path is impossible to lead to a solution
        if abs(difference) > (rows - 1 - row_index) + (columns - 1 - column_index):
            return False

        # Update the difference based on the current cell value.
        current_difference = difference + (1 if grid[row_index][column_index] == 1 else -1)

        # Explore the path downwards, if possible.
        if row_index + 1 < rows:
            if depth_first_search(row_index + 1, column_index, current_difference):
                return True

        # Explore the path to the right, if possible.
        if column_index + 1 < columns:
            if depth_first_search(row_index, column_index + 1, current_difference):
                return True

        return False

    # Initiate the depth-first search from the top-left corner.
    return depth_first_search(0, 0, 0)

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm explores all possible paths from the top-left to the bottom-right of the m x n grid using a recursive or iterative approach (e.g., Depth-First Search or Breadth-First Search). In the worst-case scenario, it might have to visit each cell in the grid. The number of possible paths can be significant. Since the algorithm visits each cell at most once and performs a constant amount of work (updating the counter and checking the conditions) at each cell, the time complexity is proportional to the number of cells in the grid, which is m*n. Therefore, the overall time complexity is O(m*n) where m and n are dimensions of the grid.

Space Complexity

O(M*N*K) – The space complexity is primarily determined by the data structure used to keep track of visited counter values along the paths. In the worst-case scenario, we might need to store counter values for a significant portion of the grid's possible paths. Let M be the number of rows, N be the number of columns, and K be the range of possible counter values (1s - 0s). Therefore, the space required could grow proportionally to M*N*K, where we store the visited states to determine equal paths.

Edge Cases

Case	How to Handle
Null or Empty Matrix	Return false immediately if the input matrix is null or empty as no path exists.
1x1 Matrix	Return true if the single element is 0, otherwise return false as a path of length 1 cannot have an equal number of 0s and 1s.
Matrix with only 0s	Return false if matrix size is odd; otherwise, path possible if dimensions are the same (square matrix of even dimension)
Matrix with only 1s	Return false if matrix size is odd; otherwise, path possible if dimensions are the same (square matrix of even dimension)
Rectangular Matrix with uneven row and column lengths	The algorithm should correctly traverse and check the counts regardless of the matrix's rectangular shape.
Large Matrix (potential StackOverflowError or Time Limit Exceeded)	Use dynamic programming or iterative approach instead of recursion to avoid stack overflow and improve efficiency.
No path exists with equal 0s and 1s	The algorithm should return false if it explores all possible paths and finds none satisfying the condition.
Integer Overflow in path length or count	Use appropriate data types (e.g., long) for counts to avoid integer overflow if matrix dimensions are very large.

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, and what is the maximum possible size for the grid's dimensions?
What values can the grid cells contain besides 0 and 1? Are negative numbers or null values possible?
Can I only move right and down in the grid, or are other movement directions allowed?
If no path exists with an equal number of 0's and 1's, what should the function return? Should I return null, false, or throw an exception?
Is there any limit on the number of steps/cells I can take in the path, or is the only restriction to reach the destination?

Brute Force Solution

Approach

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

Start at the beginning of the grid.
Explore every possible path from the starting point to the end of the grid.
For each path, keep track of how many zeros and how many ones you see along the way.
After you reach the end of the grid for a particular path, check if the number of zeros you counted is equal to the number of ones you counted.
If they are equal, you've found a valid path.
Continue exploring all other possible paths, repeating the counting and checking steps.
If any path has an equal number of zeros and ones, you can say that a valid path exists.

Code Implementation

def path_with_equal_zeros_and_ones_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    def explore_path(row_index, column_index, current_path_zeros, current_path_ones):
        #If we reach the end, check if 0s equal 1s
        if row_index == number_of_rows - 1 and column_index == number_of_columns - 1:
            if matrix[row_index][column_index] == 0:
                current_path_zeros += 1
            else:
                current_path_ones += 1
            return current_path_zeros == current_path_ones

        if matrix[row_index][column_index] == 0:
            current_path_zeros += 1
        else:
            current_path_ones += 1

        # Explore right and down if possible
        path_exists = False
        if row_index + 1 < number_of_rows:
            path_exists = path_exists or explore_path(row_index + 1, column_index, current_path_zeros, current_path_ones)

        if column_index + 1 < number_of_columns:
            path_exists = path_exists or explore_path(row_index, column_index + 1, current_path_zeros, current_path_ones)

        return path_exists

    #Start at the top left (0,0)
    return explore_path(0, 0, 0, 0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the top-left corner of the grid and initialize a counter to zero. This counter represents the difference between 1s and 0s.
As you move from one cell to the next, either down or to the right, update the counter. If the cell contains a '1', increase the counter; if it contains a '0', decrease it.
Keep track of all the counter values you've seen along the way. If you ever revisit a counter value, it means you can reach that point with multiple different paths having the same difference between 1s and 0s.
When you reach the bottom-right corner, check if the counter is zero. A zero counter means you've found a path where the number of 1s and 0s are equal.
To make it efficient, we can stop exploring a path if it's impossible to reach zero at the end. We can do this by checking if remaining steps can compensate for current difference between the 1s and 0s.

Code Implementation

def has_valid_path(grid):
    rows = len(grid)
    columns = len(grid[0])

    def depth_first_search(row_index, column_index, difference):
        # Base case: Reached the destination.
        if row_index == rows - 1 and column_index == columns - 1:
            return difference + (1 if grid[row_index][column_index] == 1 else -1) == 0

        # Pruning: Check if the path is impossible to lead to a solution
        if abs(difference) > (rows - 1 - row_index) + (columns - 1 - column_index):
            return False

        # Update the difference based on the current cell value.
        current_difference = difference + (1 if grid[row_index][column_index] == 1 else -1)

        # Explore the path downwards, if possible.
        if row_index + 1 < rows:
            if depth_first_search(row_index + 1, column_index, current_difference):
                return True

        # Explore the path to the right, if possible.
        if column_index + 1 < columns:
            if depth_first_search(row_index, column_index + 1, current_difference):
                return True

        return False

    # Initiate the depth-first search from the top-left corner.
    return depth_first_search(0, 0, 0)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or Empty Matrix	Return false immediately if the input matrix is null or empty as no path exists.
1x1 Matrix	Return true if the single element is 0, otherwise return false as a path of length 1 cannot have an equal number of 0s and 1s.
Matrix with only 0s	Return false if matrix size is odd; otherwise, path possible if dimensions are the same (square matrix of even dimension)
Matrix with only 1s	Return false if matrix size is odd; otherwise, path possible if dimensions are the same (square matrix of even dimension)
Rectangular Matrix with uneven row and column lengths	The algorithm should correctly traverse and check the counts regardless of the matrix's rectangular shape.
Large Matrix (potential StackOverflowError or Time Limit Exceeded)	Use dynamic programming or iterative approach instead of recursion to avoid stack overflow and improve efficiency.
No path exists with equal 0s and 1s	The algorithm should return false if it explores all possible paths and finds none satisfying the condition.
Integer Overflow in path length or count	Use appropriate data types (e.g., long) for counts to avoid integer overflow if matrix dimensions are very large.