Problem Description

The problem provides a representation of a bank's floor plan using a binary string array bank, where each string represents a row in the bank. A '0' indicates an empty cell and a '1' indicates the presence of a security device. Laser beams are formed between security devices on different rows, but only if there are no security devices on the rows between them. Your task is to determine the total number of laser beams in the bank.

To better understand the problem, consider the following points:

• The floor plan is visualized as a grid, where each row is a string from the array, and each character in the string corresponds to a cell in that row.
• A laser beam can only form vertically between devices in columns where there happens to be a device in both rows with no devices in the intervening rows.
• Multiple beams can form in the same column if there are more than two devices on that column but separated by rows without devices.

The objective is to calculate the total number of these beams. Imagine stacking the rows on top of each other: a laser beam is a vertical line connecting '1's but only if the slice of column between them has no other '1's.

Intuition

To come up with the solution, we must track security devices along each row and identify potential beam formations. The approach is as follows:

• Go through each row of the bank and count the number of security devices ('1's).
• Keep track of the number of devices from the last row that had devices. Initially, this is zero since we haven't encountered any devices.
• As we continue to check each row, when we find a row with devices, we can potentially form laser beams with the devices from the previously encountered row. The number of new beams formed is the product of the number of devices in the current row and the number of devices in the previously encountered row.
• We then update the 'last' variable to hold the count of devices in the current row, as this will be used to calculate beams with the next row that contains devices.

Following this logic, the algorithm tallies the number of beams as we iterate over the rows. This systematic approach ensures that all possible beam connections are counted without missing any viable combinations or double-counting beams.

Learn more about Math patterns.

Solution Approach

The given Python implementation begins by initializing two variables: last to hold the count of devices from the last row that contained devices, and ans to accumulate the total number of laser beams.

The code then iterates over each row (b) in the bank:

for b in bank:
    ...

For each row, the code uses the count method to determine the number of '1's in it which translates to the number of devices in that row:

if (t := b.count('1')) > 0:

This count is assigned to t, and the use of the walrus operator (:=) helps condense the assignment and the condition into one line.

When the number of devices t is greater than zero, meaning the current row has devices, two things happen:

1. The code calculates the number of laser beams formed with the previous row by multiplying last (the count from the last row with devices) by t (the count from the current row):

ans += last * t

2. It then updates the last variable to t, so that it holds the current count for the next iteration:

last = t

The rationale behind this is simple: laser beams only form between devices from different rows. Thus, the current row can only form beams with a previous row, not with itself or rows it hasn't 'seen' yet.

After iterating through all the rows, ans holds the total number of beams, and the code returns this value:

return ans

The algorithm's complexity is O(m * n), where m is the number of rows and n is the number of columns (or the length of each row), as it needs to count the '1's in each row.

This solution is efficient and avoids unnecessary calculations by not looking at the individual positions of devices within the rows. It leverages the fact that any two devices in the same column from different rows would form beams with all pairs of rows between them that don't contain any devices.

Example Walkthrough

Let us consider a small example to illustrate the solution approach.

Suppose we have the following bank array representing our bank's floor plan:

["011",
 "000",
 "001"]

This array represents a bank with three rows and three columns. Each '1' is a security device, and each '0' is an empty space.

Using the intuition and solution approach:

1. We initialize last to 0 (since we have not processed any rows yet) and ans to 0 (as our accumulator for laser beams).

2. We start iterating through the rows:

   • For the first row ("011"), there are 2 security devices. With last initially being 0, no beams can be formed yet, so we simply set last to 2.

   • For the second row ("000"), there are no security devices. Nothing happens here since we need at least one security device to form a laser beam.

   • For the third row ("001"), there is 1 security device. We find the product of last (which is 2 from the first row) and the count of devices in the current row (1), resulting in 2 laser beams formed. We add this to ans, giving us 0 + (2 * 1) = 2 beams.

   • We update last to 1, reflecting the number of devices in the third row.

3. There are no more rows, so we conclude the example with ans being 2. There are 2 laser beams formed between the devices in the first and third rows.

Thus the final output for this example bank is 2 laser beams.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is O(n*m), where n is the number of rows in the bank list and m is the number of columns in each row. This is because the code iterates over each row with a loop (which runs n times), and within each iteration, it performs a count operation with b.count('1'), which, in the worst-case scenario, runs in O(m) time for each row if all elements need to be checked (where m can be seen as the maximum possible length of the binary strings representing the devices). The other operations inside the loop are constant time operations, therefore, they do not change the overall time complexity.

Space Complexity

The space complexity of the code is O(1). The solution uses a constant amount of extra space regardless of the input size. Variables last, ans, and t are reused for each row, hence, the space used does not grow with the size of the input (bank list).

Learn more about how to find time and space complexity quickly using problem constraints.