Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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:
r1 and r2, where r1 < r2.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.lengthn == bank[i].length1 <= m, n <= 500bank[i][j] is either '0' or '1'.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math
["011001","000000","010100","001000"]
["000","111","000"]
set-matrix-zeroes)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2125: Number of Laser Beams in a Bank
class Solution {
public int numberOfBeams(String[] bank) {
int ans = 0, pre = 0;
for (String row : bank) {
int cur = 0;
for (int i = 0; i < row.length(); ++i) {
cur += row.charAt(i) - '0';
}
if (cur > 0) {
ans += pre * cur;
pre = cur;
}
}
return ans;
}
}
// Accepted solution for LeetCode #2125: Number of Laser Beams in a Bank
func numberOfBeams(bank []string) (ans int) {
pre := 0
for _, row := range bank {
if cur := strings.Count(row, "1"); cur > 0 {
ans += pre * cur
pre = cur
}
}
return
}
# Accepted solution for LeetCode #2125: Number of Laser Beams in a Bank
class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
ans = pre = 0
for row in bank:
if (cur := row.count("1")) > 0:
ans += pre * cur
pre = cur
return ans
// Accepted solution for LeetCode #2125: Number of Laser Beams in a Bank
impl Solution {
pub fn number_of_beams(bank: Vec<String>) -> i32 {
let mut ans = 0;
let mut pre = 0;
for row in bank {
let cur = row.chars().filter(|&c| c == '1').count() as i32;
if cur > 0 {
ans += pre * cur;
pre = cur;
}
}
ans
}
}
// Accepted solution for LeetCode #2125: Number of Laser Beams in a Bank
function numberOfBeams(bank: string[]): number {
let [ans, pre] = [0, 0];
for (const row of bank) {
const cur = row.split('1').length - 1;
if (cur) {
ans += pre * cur;
pre = cur;
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.