# Solving Leetcode Interviews in Seconds with AI: Search a 2D Matrix II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "240" 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
	> Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:  Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.    Example 1:   Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5 Output: true  Example 2:   Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20 Output: false    Constraints:  m == matrix.length n == matrix[i].length 1 <= n, m <= 300 -109 <= matrix[i][j] <= 109 All the integers in each row are sorted in ascending order. All the integers in each column are sorted in ascending order. -109 <= target <= 109  

	# Explanation
	Here's a solution to efficiently search for a target value in a sorted matrix with the given properties:

*   **Start at the top-right corner:** Begin the search from the element at `matrix[0][n-1]`. This position allows us to simultaneously eliminate either a row or a column in each comparison.
*   **Comparison and Movement:**  Compare the current element with the target.
    *   If the current element is equal to the target, return `true`.
    *   If the current element is greater than the target, it means all elements in the current column below this element are also greater than the target (due to column-wise sorting). So, move one column to the left.
    *   If the current element is smaller than the target, it means all elements in the current row to the left of this element are also smaller than the target (due to row-wise sorting). So, move one row down.
*   **Out-of-bounds Check:**  Continue these comparisons and movements until either the target is found or the search goes out of bounds (i.e., the row index becomes greater than or equal to `m`, or the column index becomes less than 0). If the target is not found after the search completes, return `false`.

*   **Runtime Complexity:** O(m + n), where 'm' is the number of rows and 'n' is the number of columns.
*   **Storage Complexity:** O(1) (constant).

	
	# Code
	```python
	def searchMatrix(matrix, target):
    """
    Searches for a target value in a sorted matrix.

    Args:
        matrix: A list of lists of integers, where each row and column is sorted in ascending order.
        target: The integer to search for.

    Returns:
        True if the target is found in the matrix, False otherwise.
    """
    if not matrix or not matrix[0]:
        return False

    m = len(matrix)
    n = len(matrix[0])
    row = 0
    col = n - 1

    while row < m and col >= 0:
        if matrix[row][col] == target:
            return True
        elif matrix[row][col] > target:
            col -= 1
        else:
            row += 1

    return False
	```