# Solving Leetcode Interviews in Seconds with AI: Equal Row and Column Pairs


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2352" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal. A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).   Example 1:   Input: grid = [[3,2,1],[1,7,6],[2,7,7]] Output: 1 Explanation: There is 1 equal row and column pair: - (Row 2, Column 1): [2,7,7]  Example 2:   Input: grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]] Output: 3 Explanation: There are 3 equal row and column pairs: - (Row 0, Column 0): [3,1,2,2] - (Row 2, Column 2): [2,4,2,2] - (Row 3, Column 2): [2,4,2,2]    Constraints:  n == grid.length == grid[i].length 1 <= n <= 200 1 <= grid[i][j] <= 105  

	# Explanation
	*   **Hashing:** Convert each row and column into a string or tuple representation and use a hash map (dictionary) to store the frequency of each row.
*   **Comparison:** Iterate through the columns, convert each column to its string/tuple representation, and check if it exists in the row frequency map. If it does, increment the count by the frequency of that row.
*   **Return Count:** The final count will be the number of equal row and column pairs.

*   **Runtime Complexity:** O(n^2), **Storage Complexity:** O(n)

	
	# Code
	```python
	def equalPairs(grid):
    n = len(grid)
    row_counts = {}
    count = 0

    # Convert rows to tuples and count their occurrences
    for row in grid:
        row_tuple = tuple(row)
        row_counts[row_tuple] = row_counts.get(row_tuple, 0) + 1

    # Iterate through columns and check for matches
    for j in range(n):
        column = []
        for i in range(n):
            column.append(grid[i][j])
        col_tuple = tuple(column)

        if col_tuple in row_counts:
            count += row_counts[col_tuple]

    return count
	```