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Equal Row and Column Pairs

Medium

Equal Row and Column Pairs

Medium

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 is the maximum size of the input matrix (n)?
Can the matrix contain negative integers or only positive integers?
If no equal row and column pairs exist, should I return 0?
Are the rows and columns considered equal only if their elements are in the exact same order?
Does the matrix always have to be square (n x n), or can it be rectangular (n x m)?

Brute Force Solution

Approach

The brute force approach to finding equal row and column pairs is like meticulously comparing every row with every column. We will go through each possible row and column combination, checking if they are identical. This method guarantees we find all matching pairs.

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

Take the first row of the entire grid.
Compare this row to the first column of the entire grid.
If they are exactly the same, count it as a match.
Next, compare the same row to the second column.
Repeat this process, comparing the first row to every column in the grid.
Once you have compared the first row to all the columns, move on to the second row.
Compare the second row to every column in the grid, one column at a time, just like before, and keep track of any matches.
Continue this process, comparing each row to every column, until you have gone through all the rows.
The final count of matches is the total number of equal row and column pairs.

Code Implementation

def equal_row_and_column_pairs(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])
    equal_pairs_count = 0

    # Iterate through each row of the grid
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            # Compare the current row with the current column
            row_elements = grid[row_index]
            column_elements = []
            for row_num in range(number_of_rows):
                column_elements.append(grid[row_num][column_index])

            # Check if the row and column are equal
            if row_elements == column_elements:
                # Increment the counter if row and column match
                equal_pairs_count += 1

    return equal_pairs_count

Big(O) Analysis

Time Complexity

O(n*n*n) – The algorithm iterates through each of the n rows in the grid. For each row, it iterates through each of the n columns in the grid. Within the nested loops, it compares the row with the column element by element, which takes O(n) time in the worst case. Therefore, the overall time complexity is O(n * n * n), which simplifies to O(n³).

Space Complexity

O(1) – The brute force approach, as described, iterates through rows and columns comparing them directly. No auxiliary data structures like temporary arrays, lists, or hash maps are created to store intermediate results or track visited elements. The algorithm only uses a few index variables for iterating through the grid, and the space required for these variables remains constant regardless of the input grid size N. Therefore, the space complexity is O(1), constant space.

Optimal Solution

Approach

The problem asks us to find matching row and column pairs in a grid. The clever approach is to count how many times each row appears and how many times each column appears, then efficiently find the overlaps between these counts.

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

First, let's convert each row in the grid into a simple description, like a string. This helps us easily compare rows later.
Next, count how many times each of these row descriptions shows up in the grid. This tells us how many identical rows we have of each type.
Then, do the same thing for the columns: create descriptions and count the frequency of each column.
Now, go through each row description and see if its description also appears among the column descriptions. If it does, this means we have at least one matching pair.
Multiply the number of times the row description appears by the number of times the matching column description appears. This gives us the number of pairs we can make with that specific row and column.
Add up all these products for each row. The final sum is the total number of equal row and column pairs in the grid.

Code Implementation

def equal_row_column_pairs(grid):
    number_of_rows = len(grid)
    number_of_cols = len(grid[0])

    row_counts = {}
    for row in grid:
        row_tuple = tuple(row)
        row_counts[row_tuple] = row_counts.get(row_tuple, 0) + 1

    col_counts = {}
    for column_index in range(number_of_cols):
        column = []
        for row_index in range(number_of_rows):
            column.append(grid[row_index][column_index])
        column_tuple = tuple(column)
        col_counts[column_tuple] = col_counts.get(column_tuple, 0) + 1

    equal_pairs_count = 0
    # Iterate through each row to find matching columns.
    for row_tuple, row_frequency in row_counts.items():
        # If a row exists as a column, compute number of pairs
        if row_tuple in col_counts:
            equal_pairs_count += row_frequency * col_counts[row_tuple]

    # Sum up the number of equal row and column pairs.
    return equal_pairs_count

Big(O) Analysis

Time Complexity

O(n²) – Converting each row and column into a string representation takes O(n) time, and this is done for each of the n rows and n columns, resulting in O(n*n) = O(n²) time. Counting the frequency of each row and column using a hash map takes O(n) time, as we iterate through n rows and n columns respectively and perform constant time lookups and insertions. Finally, iterating through the row frequencies and checking for matching column frequencies also takes O(n) time. Because the dominant factor is converting rows and columns to a string and then iterating over them, the overall time complexity approximates O(n*n) and simplifies to O(n²).

Space Complexity

O(N^2) – The algorithm uses hash maps (or dictionaries) to store row and column descriptions and their frequencies. In the worst case, all rows and all columns are unique. Creating these descriptions as strings requires storing N strings of length N (assuming the grid is NxN), which leads to O(N^2) space. Additionally, the hash maps themselves store these N string keys along with their counts, also contributing to O(N^2) space. Therefore, the overall auxiliary space complexity is O(N^2).

Edge Cases

Case	How to Handle
Null or empty input matrix	Return 0 since there are no row/column pairs to count.
Square matrix with only one row/column	Check if the single row is equal to the single column and return 1 if it is, 0 otherwise.
Large matrix with dimensions near the maximum integer value.	Ensure the chosen data structures (e.g., hash maps) can accommodate the potential number of rows/columns without memory issues.
Matrix where all rows are identical and all columns are identical	The algorithm should correctly count all pairs, potentially leading to a quadratic number of pairs if row and column counts are high, so ensure efficient calculation.
Matrix with negative numbers, zeros, and large positive numbers	Ensure numerical comparisons handle the full range of possible integer values without integer overflow.
Matrix where no row is equal to any column.	The algorithm should correctly return 0 in this case, indicating no matching pairs found.
Input matrix is not a square matrix	The algorithm needs to correctly iterate through all rows and columns regardless of whether the matrix is square or rectangular.
Rows/columns are equal but contain data types that do not match precisely (e.g., strings vs. numbers)	The problem constraints suggest numerical comparison, but if the input data types are mixed, cast all to the expected type before comparison or throw an error.

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 is the maximum size of the input matrix (n)?
Can the matrix contain negative integers or only positive integers?
If no equal row and column pairs exist, should I return 0?
Are the rows and columns considered equal only if their elements are in the exact same order?
Does the matrix always have to be square (n x n), or can it be rectangular (n x m)?

Brute Force Solution

Approach

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

Take the first row of the entire grid.
Compare this row to the first column of the entire grid.
If they are exactly the same, count it as a match.
Next, compare the same row to the second column.
Repeat this process, comparing the first row to every column in the grid.
Once you have compared the first row to all the columns, move on to the second row.
Compare the second row to every column in the grid, one column at a time, just like before, and keep track of any matches.
Continue this process, comparing each row to every column, until you have gone through all the rows.
The final count of matches is the total number of equal row and column pairs.

Code Implementation

def equal_row_and_column_pairs(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])
    equal_pairs_count = 0

    # Iterate through each row of the grid
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            # Compare the current row with the current column
            row_elements = grid[row_index]
            column_elements = []
            for row_num in range(number_of_rows):
                column_elements.append(grid[row_num][column_index])

            # Check if the row and column are equal
            if row_elements == column_elements:
                # Increment the counter if row and column match
                equal_pairs_count += 1

    return equal_pairs_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, let's convert each row in the grid into a simple description, like a string. This helps us easily compare rows later.
Next, count how many times each of these row descriptions shows up in the grid. This tells us how many identical rows we have of each type.
Then, do the same thing for the columns: create descriptions and count the frequency of each column.
Now, go through each row description and see if its description also appears among the column descriptions. If it does, this means we have at least one matching pair.
Multiply the number of times the row description appears by the number of times the matching column description appears. This gives us the number of pairs we can make with that specific row and column.
Add up all these products for each row. The final sum is the total number of equal row and column pairs in the grid.

Code Implementation

def equal_row_column_pairs(grid):
    number_of_rows = len(grid)
    number_of_cols = len(grid[0])

    row_counts = {}
    for row in grid:
        row_tuple = tuple(row)
        row_counts[row_tuple] = row_counts.get(row_tuple, 0) + 1

    col_counts = {}
    for column_index in range(number_of_cols):
        column = []
        for row_index in range(number_of_rows):
            column.append(grid[row_index][column_index])
        column_tuple = tuple(column)
        col_counts[column_tuple] = col_counts.get(column_tuple, 0) + 1

    equal_pairs_count = 0
    # Iterate through each row to find matching columns.
    for row_tuple, row_frequency in row_counts.items():
        # If a row exists as a column, compute number of pairs
        if row_tuple in col_counts:
            equal_pairs_count += row_frequency * col_counts[row_tuple]

    # Sum up the number of equal row and column pairs.
    return equal_pairs_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input matrix	Return 0 since there are no row/column pairs to count.
Square matrix with only one row/column	Check if the single row is equal to the single column and return 1 if it is, 0 otherwise.
Large matrix with dimensions near the maximum integer value.	Ensure the chosen data structures (e.g., hash maps) can accommodate the potential number of rows/columns without memory issues.
Matrix where all rows are identical and all columns are identical	The algorithm should correctly count all pairs, potentially leading to a quadratic number of pairs if row and column counts are high, so ensure efficient calculation.
Matrix with negative numbers, zeros, and large positive numbers	Ensure numerical comparisons handle the full range of possible integer values without integer overflow.
Matrix where no row is equal to any column.	The algorithm should correctly return 0 in this case, indicating no matching pairs found.
Input matrix is not a square matrix	The algorithm needs to correctly iterate through all rows and columns regardless of whether the matrix is square or rectangular.
Rows/columns are equal but contain data types that do not match precisely (e.g., strings vs. numbers)	The problem constraints suggest numerical comparison, but if the input data types are mixed, cast all to the expected type before comparison or throw an error.