2352. Equal Row and Column Pairs

Problem Description

The problem presents a square integer matrix (grid) that has a size of n x n. We are tasked with finding the number of pairs (r_i, c_j), where r_i represents the rows and c_j represents the columns of the matrix. For a pair of row and column to be counted, they must have the same elements in the exact same order - essentially, they should form identical arrays.

To better understand the problem, let's consider a simple example. Suppose we have the following matrix:

1grid = [
2  [1, 2],
3  [2, 1]
4]

Here, we can compare each row with each column. We can see that the first row [1, 2] does not match either column (first column is [1, 2], second column is [2, 1]). However, the second row [2, 1] matches the second column [2, 1]. Thus, there is only one equal pair in this matrix: the pair consisting of the second row and the second column.

Intuition

When approaching this solution, we have to consider that we can't simply compare each row directly to each column since they are not stored in the same format within the matrix. Rows are sublists of the matrix, while columns can be viewed as tuples formed by taking the same element index from each row.

The solution employs two main steps:

1. Transpose the matrix: This is achieved using the zip function along with unpacking the original grid matrix. By doing this, we obtain a transposed version of the original grid, where rows are now columns and vice versa. For the given example, the transposed grid would look like this:

   1g = [
   2  [1, 2],
   3  [2, 1]
   4]

2. Count the equal pairs: We can now iterate over each row of the original grid and for each row, iterate over each row of the transposed grid (g). Whenever we find a row in the original grid matches a row in g, it means that the original row and the corresponding column of the original grid form an equal pair. We sum up all such instances to get the final result.

The intuition is that the transposed version of the grid will make it easier to compare rows and columns, and using a simple nested iteration, we can then count how many such comparisons yield identical arrays which indicate the pairs of rows and columns that are equal.

Solution Approach

The implementation of the solution can be broken down as follows:

• The zip function is a built-in Python function used to aggregate elements from two or more iterables (lists, tuples, etc.). It takes n iterables and returns a list (or in Python 3.x, an iterator) of tuples, where the i-th tuple contains the i-th element from each of the iterable arguments. Hence, applying zip to the original grid matrix, while using the unpacking operator *, effectively transposes the matrix. This new matrix g holds the columns of the original grid as its rows.

  The best way to visualize this is by comparison:

  Original grid:

  1[
  2  [1, 2],
  3  [3, 4]
  4]

  Transposed g using zip:

  1[
  2  [1, 3],
  3  [2, 4]
  4]

• The next step is using list comprehension to iterate through each row of the original grid and simultaneously through each row of the transposed grid g. At every iteration, the row from the original grid is compared to the current row in g using the == operator, which checks if both lists contain the same elements in the same order.

• The condition row == col returns a True (which is interpreted as 1) if the row and column are identical, and False (interpreted as 0) if they aren't. The sum function is then used to add up these truth values to get the total count of matching pairs.

• The technique of using a nested for loop within the list comprehension is a common algorithm pattern that allows for checking each combination of elements between two sequences or structures. This pattern is known as "Cartesian product" and is essential in different domains for creating every possible pairing of sets of items.

Here is how the Python code uses these steps with the algorithmic patterns:

1class Solution:
2    def equalPairs(self, grid: List[List[int]]) -> int:
3        g = [list(col) for col in zip(*grid)]  # Transpose the grid by converting the zipped tuples to lists
4        return sum(row == col for row in grid for col in g)  # Count equal row-column pairs

As we can see, the solution is concise and leverages powerful built-in Python functions and list comprehensions to solve the problem efficiently. The time complexity of this solution is O(n^2), where n is the size of the grid's row or column since every row is compared with every column.

Example Walkthrough

Consider a 3x3 square matrix as our grid:

1grid = [
2  [8, 2, 9],
3  [5, 1, 4],
4  [7, 6, 3]
5]

To find the number of equal pairs of rows and columns, let's apply our solution approach step by step:

1. Transpose the matrix grid:

By using the zip function with unpacking, we transpose the grid. The transposed matrix g would be the following:

1g = [
2  [8, 5, 7],
3  [2, 1, 6],
4  [9, 4, 3]
5]

This is the visual representation of how the zip function has taken the first element of each row and created the first row of g, and so on.

2. Count the equal pairs:

Now we compare each row of the original grid with each row of the transposed g and count the instances where they are identical:

• Compare grid row [8, 2, 9] with g:

  • It does not match any row in g.

• Compare grid row [5, 1, 4] with g:

  • It matches the first row of g ([8, 5, 7]) at the second position, but each row has different elements.

• Compare grid row [7, 6, 3] with g:

  • The third row of grid ([7, 6, 3]) is identical to the third row of g ([7, 6, 3]), so we have a match.

In this case, we find that there is exactly one equal pair consisting of the third row of grid and the third row (which corresponds to the third column of the original grid) of g.

Using the code provided in the solution approach, we would have:

1class Solution:
2    def equalPairs(self, grid: List[List[int]]) -> int:
3        g = [list(col) for col in zip(*grid)] # This makes `g` as shown above
4        return sum(row == col for row in grid for col in g) # This counts and returns 1 for our example

When run with our example grid, equalPairs would return 1, indicating that there is one pair of a row and a column that has the same elements in the exact same order.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n^2) where n is the size of one dimension of the square matrix grid. This is because the code includes two nested loops each iterating through n elements (rows and columns). The comparison row == col inside the loops is executed n * n times. Each comparison operation takes O(n), thus overall, the time complexity is O(n^3).

The space complexity of the code is O(n^2). The list g is created by taking a transpose of the original grid, which requires additional space proportional to the size of the grid, hence n * n.

Learn more about how to find time and space complexity quickly using problem constraints.