LeetCode Challenge Day 42 — 2125. Number of Laser Beams in a Bank

Hey folks

This is Day 42 of my LeetCode streak 🚀.
Today’s problem is 2125. Number of Laser Beams in a Bank — a matrix simulation problem where we count beams formed by security devices placed in rows.

📌 Problem Statement

You are given a binary string array bank representing rows of a bank security floor.

• '1' represents a security device.
• '0' means no device.

A laser beam is formed between two devices in different rows if and only if all rows between them are empty (contain no devices).

Return the total number of beams.

💡 Intuition

Beams only occur between consecutive non-empty rows.
So we can:

1. Count the number of devices in each row.
2. For every non-empty row, multiply its count with the previous non-empty row’s count.
3. Accumulate the result.

🔑 Approach

1. Initialize prev = 0 (count of devices in last non-empty row) and beams = 0.
2. Traverse each row:
   • Count devices (curr).
   • If curr > 0:
     • Add prev * curr to beams.
     • Update prev = curr.
   • Otherwise skip (empty row).
3. Return beams.

⏱️ Complexity Analysis

• Time complexity:
  We scan each row of length n for m rows → O(m · n).

• Space complexity:
  Only a few variables (prev, curr, beams) → O(1).

🧑‍💻 Code (JavaScript)

/**
 * @param {string[]} bank
 * @return {number}
 */
var numberOfBeams = function (bank) {
  let prev = 0;     // number of devices in the previous non-empty row
  let beams = 0;

  for (const row of bank) {
    // count '1's in this row
    let curr = 0;
    for (let i = 0; i < row.length; i++) {
      if (row[i] === '1') curr++;
    }

    // if this row has devices, it forms beams with the previous non-empty row
    if (curr > 0) {
      beams += prev * curr;
      prev = curr; // update previous non-empty row count
    }
  }

  return beams;
};

🧪 Example Walkthrough

Input: bank = ["011001","000000","010100","001000"]

• Row 1 → 3 devices.
• Row 2 → empty (skip).
• Row 3 → 2 devices → beams += 3 * 2 = 6.
• Row 4 → 1 device → beams += 2 * 1 = 2.

Total = 8 ✅

🎥 Reflections

This problem highlights the importance of simplifying conditions. Instead of checking all row pairs, tracking only consecutive non-empty rows gives a clean and efficient solution.

That’s it for Day 42 of my LeetCode journey! Onwards to the next challenge 🔥

Happy Coding 👨‍💻