Introduction

In this blog post, we will explore how to solve the LeetCode problem "2125" 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

Anti-theft security devices are activated inside a bank. You are given a 0-indexed binary string array bank representing the floor plan of the bank, which is an m x n 2D matrix. bank[i] represents the ith row, consisting of '0's and '1's. '0' means the cell is empty, while'1' means the cell has a security device. There is one laser beam between any two security devices if both conditions are met: The two devices are located on two different rows: r1 and r2, where r1 < r2. For each row i where r1 < i < r2, there are no security devices in the ith row. Laser beams are independent, i.e., one beam does not interfere nor join with another. Return the total number of laser beams in the bank. Example 1: Input: bank = ["011001","000000","010100","001000"] Output: 8 Explanation: Between each of the following device pairs, there is one beam. In total, there are 8 beams: bank[0][1] -- bank[2][1] bank[0][1] -- bank[2][3] bank[0][2] -- bank[2][1] bank[0][2] -- bank[2][3] bank[0][5] -- bank[2][1] bank[0][5] -- bank[2][3] bank[2][1] -- bank[3][2] bank[2][3] -- bank[3][2] Note that there is no beam between any device on the 0th row with any on the 3rd row. This is because the 2nd row contains security devices, which breaks the second condition. Example 2: Input: bank = ["000","111","000"] Output: 0 Explanation: There does not exist two devices located on two different rows. Constraints: m == bank.length n == bank[i].length 1 <= m, n <= 500 bank[i][j] is either '0' or '1'.

Explanation

• Iterate through the rows of the bank. Count the number of security devices ('1's) in each row.
  • Keep track of the previous row's security device count. If a row has security devices, calculate the number of laser beams between the current row and the previous row with security devices.
  • Sum the number of laser beams calculated in each valid row pair.

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

Code

    def numberOfBeams(bank):
    """
    Calculates the total number of laser beams in the bank.

    Args:
        bank: A 0-indexed binary string array representing the floor plan of the bank.

    Returns:
        The total number of laser beams in the bank.
    """

    total_beams = 0
    previous_devices = 0

    for row in bank:
        current_devices = row.count('1')

        if current_devices > 0:
            if previous_devices > 0:
                total_beams += previous_devices * current_devices
            previous_devices = current_devices

    return total_beams