Number of Laser Beams in a Bank

Tested Python solution for LeetCode 2125 with 32 pytest cases. Generate a practice environment with lcpy.

LeetCode 2125, Medium. Topics: Array, Math, String, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 32 parametrized pytest cases, and a playground notebook:

lcpy gen -n 2125   # by problem number
lcpy gen -s number_of_laser_beams_in_a_bank   # by problem name

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Problem

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.

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Examples

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.

Input: bank = ["000","111","000"]
Output: 0

Explanation: There does not exist two devices located on two different rows.

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Constraints

• m == bank.length
• n == bank[i].length
• 1 <= m, n <= 500
• bank[i][j] is either ‘0’ or ‘1’.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n)
    # Space: O(1)
    def number_of_beams(self, bank: list[str]) -> int:
        total = 0
        prev = 0
        for row in bank:
            count = row.count("1")
            if count:
                total += prev * count
                prev = count
        return total

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Complexity

Time	Space
O(m * n)	O(1)

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Tags

NeetCode All.