# Solving Leetcode Interviews in Seconds with AI: Check if Every Row and Column Contains All Numbers


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2133" 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
	> An n x n matrix is valid if every row and every column contains all the integers from 1 to n (inclusive). Given an n x n integer matrix matrix, return true if the matrix is valid. Otherwise, return false.   Example 1:   Input: matrix = [[1,2,3],[3,1,2],[2,3,1]] Output: true Explanation: In this case, n = 3, and every row and column contains the numbers 1, 2, and 3. Hence, we return true.  Example 2:   Input: matrix = [[1,1,1],[1,2,3],[1,2,3]] Output: false Explanation: In this case, n = 3, but the first row and the first column do not contain the numbers 2 or 3. Hence, we return false.    Constraints:  n == matrix.length == matrix[i].length 1 <= n <= 100 1 <= matrix[i][j] <= n  

	# Explanation
	Here's the breakdown:

*   **Key Idea:** Use sets to efficiently check if each row and each column contains all numbers from 1 to n. Iterate through rows and columns, adding elements to a set. If the size of the set after iterating a row/column isn't `n`, or if the set contains elements outside the range `[1, n]`, the matrix is invalid.

*   **Efficiency:** The use of sets provides near-constant time complexity for checking the existence of elements, which optimizes the validation process.

*   **Complexity:**
    *   Runtime Complexity: O(n^2) - iterating through each element of the n x n matrix.
    *   Storage Complexity: O(n) -  using sets of size at most n to validate each row and column.

	
	# Code
	```python
	def check_valid_matrix(matrix: list[list[int]]) -> bool:
    """
    Checks if an n x n matrix is valid, meaning every row and column contains all integers from 1 to n.

    Args:
        matrix: The n x n integer matrix.

    Returns:
        True if the matrix is valid, False otherwise.
    """
    n = len(matrix)

    # Check rows
    for row in matrix:
        row_set = set(row)
        if len(row_set) != n:
            return False
        for num in row_set:
            if not (1 <= num <= n):
                return False

    # Check columns
    for col in range(n):
        col_set = set()
        for row in range(n):
            col_set.add(matrix[row][col])

        if len(col_set) != n:
            return False
        for num in col_set:
            if not (1 <= num <= n):
                return False

    return True
	```