# Solving Leetcode Interviews in Seconds with AI: Count Submatrices with Top-Left Element and Sum Less Than k


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3070" 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
	> You are given a 0-indexed integer matrix grid and an integer k. Return the number of submatrices that contain the top-left element of the grid, and have a sum less than or equal to k.   Example 1:   Input: grid = [[7,6,3],[6,6,1]], k = 18 Output: 4 Explanation: There are only 4 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 18. Example 2:   Input: grid = [[7,2,9],[1,5,0],[2,6,6]], k = 20 Output: 6 Explanation: There are only 6 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 20.    Constraints:  m == grid.length  n == grid[i].length 1 <= n, m <= 1000  0 <= grid[i][j] <= 1000 1 <= k <= 109  

	# Explanation
	Here's the approach:

*   **Iterate through Submatrices:** Iterate through all possible submatrices that include the top-left element. This is achieved by varying the number of rows and columns included in the submatrix.
*   **Calculate Submatrix Sum Efficiently:** For each submatrix, calculate its sum.
*   **Count Valid Submatrices:** If the sum of the submatrix is less than or equal to `k`, increment the count.

*   **Runtime Complexity:** O(m\*n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def count_submatrices(grid, k):
    """
    Counts the number of submatrices containing the top-left element with a sum less than or equal to k.

    Args:
        grid: A 0-indexed integer matrix.
        k: An integer.

    Returns:
        The number of submatrices that satisfy the condition.
    """

    m = len(grid)
    n = len(grid[0])
    count = 0

    for rows in range(1, m + 1):
        for cols in range(1, n + 1):
            submatrix_sum = 0
            for i in range(rows):
                for j in range(cols):
                    submatrix_sum += grid[i][j]

            if submatrix_sum <= k:
                count += 1

    return count
	```