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, 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(mn) time and O(mn) space, where m and n are the dimensions of the matrix.

Code

    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