# Solving Leetcode Interviews in Seconds with AI: Count Submatrices With Equal Frequency of X and Y


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3212" 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 2D character matrix grid, where grid[i][j] is either 'X', 'Y', or '.', return the number of submatrices that contain:  grid[0][0] an equal frequency of 'X' and 'Y'. at least one 'X'.    Example 1:  Input: grid = [["X","Y","."],["Y",".","."]] Output: 3 Explanation:   Example 2:  Input: grid = [["X","X"],["X","Y"]] Output: 0 Explanation: No submatrix has an equal frequency of 'X' and 'Y'.  Example 3:  Input: grid = [[".","."],[".","."]] Output: 0 Explanation: No submatrix has at least one 'X'.    Constraints:  1 <= grid.length, grid[i].length <= 1000 grid[i][j] is either 'X', 'Y', or '.'.  

	# Explanation
	Here's a breakdown of the approach, complexity, and the Python code:

*   **High-Level Approach:**
    *   Iterate through all possible submatrices defined by their top-left and bottom-right corners.
    *   For each submatrix, count the occurrences of 'X' and 'Y'.
    *   If the count of 'X' equals the count of 'Y' and the submatrix contains at least one 'X', increment the result.

*   **Complexity:**
    *   Runtime: O(m<sup>2</sup>n<sup>2</sup>), where 'm' is the number of rows and 'n' is the number of columns.  This is because we iterate through all possible submatrices, which is O(m<sup>2</sup>n<sup>2</sup>), and counting within each submatrix takes O(mn) in the worst case, but these can be combined and counted at the same time.
    *   Storage: O(1). The solution uses a constant amount of extra space.

	
	# Code
	```python
	def count_submatrices(grid):
    """
    Counts the number of submatrices in a 2D grid that contain an equal frequency of 'X' and 'Y',
    and at least one 'X'.
    """
    rows = len(grid)
    cols = len(grid[0])
    count = 0

    for r1 in range(rows):
        for c1 in range(cols):
            for r2 in range(r1, rows):
                for c2 in range(c1, cols):
                    x_count = 0
                    y_count = 0
                    has_x = False

                    for i in range(r1, r2 + 1):
                        for j in range(c1, c2 + 1):
                            if grid[i][j] == 'X':
                                x_count += 1
                                has_x = True
                            elif grid[i][j] == 'Y':
                                y_count += 1

                    if has_x and x_count == y_count:
                        count += 1

    return count
	```