# Solving Leetcode Interviews in Seconds with AI: Maximal Square


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "221" 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 an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.   Example 1:   Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]] Output: 4  Example 2:   Input: matrix = [["0","1"],["1","0"]] Output: 1  Example 3:  Input: matrix = [["0"]] Output: 0    Constraints:  m == matrix.length n == matrix[i].length 1 <= m, n <= 300 matrix[i][j] is '0' or '1'.  

	# Explanation
	Here's the solution to find the largest square of 1s in a binary matrix:

*   **Dynamic Programming:**  Build a DP table where `dp[i][j]` stores the side length of the largest square ending at `matrix[i-1][j-1]`.

*   **State Transition:** If `matrix[i-1][j-1]` is '1', `dp[i][j]` is 1 + min( `dp[i-1][j]`, `dp[i][j-1]`, `dp[i-1][j-1]` ). If it's '0', `dp[i][j]` is 0.

*   **Maximum Side:** Keep track of the maximum side length found so far and return its square.

*   **Time & Space Complexity:** O(m*n) time and O(m*n) space, where m and n are the dimensions of the matrix.

	
	# Code
	```python
	def maximalSquare(matrix):
    """
    Finds the area of the largest square containing only 1's in a binary matrix.

    Args:
        matrix: A list of lists of strings, where each string is either "0" or "1".

    Returns:
        The area of the largest square containing only 1's.
    """

    if not matrix:
        return 0

    rows = len(matrix)
    cols = len(matrix[0])
    dp = [[0] * (cols + 1) for _ in range(rows + 1)]  # DP table with padding
    max_side = 0

    for i in range(1, rows + 1):
        for j in range(1, cols + 1):
            if matrix[i - 1][j - 1] == '1':
                dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1])
                max_side = max(max_side, dp[i][j])

    return max_side * max_side
	```