Introduction

In this blog post, we will explore how to solve the LeetCode problem "832" 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 n x n binary matrix image, flip the image horizontally, then invert it, and return the resulting image. To flip an image horizontally means that each row of the image is reversed. For example, flipping [1,1,0] horizontally results in [0,1,1]. To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0. For example, inverting [0,1,1] results in [1,0,0]. Example 1: Input: image = [[1,1,0],[1,0,1],[0,0,0]] Output: [[1,0,0],[0,1,0],[1,1,1]] Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]]. Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]] Example 2: Input: image = [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]] Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]] Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]]. Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]] Constraints: n == image.length n == image[i].length 1 <= n <= 20 images[i][j] is either 0 or 1.

Explanation

• In-place reversal and inversion: Iterate through each row and reverse it in-place. While reversing, simultaneously invert each element by XORing it with 1. This avoids extra memory allocation.
  • Optimized Inversion: Use XOR operation to invert the bit without conditional statements.
  • Combined operations: Merge the reversing and inverting into a single loop for efficiency.

• Time Complexity: O(n*n), where n is the dimension of the square matrix.
• Space Complexity: O(1) - in-place operation

Code

    def flipAndInvertImage(image: list[list[int]]) -> list[list[int]]:
    """
    Flips the image horizontally and then inverts it.

    Args:
        image: A list of lists of integers representing the image.

    Returns:
        A list of lists of integers representing the flipped and inverted image.
    """
    for row in image:
        for i in range((len(row) + 1) // 2):
            row[i], row[len(row) - 1 - i] = row[len(row) - 1 - i] ^ 1, row[i] ^ 1
    return image