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.
You are given the two integers, n and m and two integer arrays, hBars and vBars. The grid has n + 2 horizontal and m + 2 vertical bars, creating 1 x 1 unit cells. The bars are indexed starting from 1.
You can remove some of the bars in hBars from horizontal bars and some of the bars in vBars from vertical bars. Note that other bars are fixed and cannot be removed.
Return an integer denoting the maximum area of a square-shaped hole in the grid, after removing some bars (possibly none).
Example 1:
Input: n = 2, m = 1, hBars = [2,3], vBars = [2]
Output: 4
Explanation:
The left image shows the initial grid formed by the bars. The horizontal bars are [1,2,3,4], and the vertical bars are [1,2,3].
One way to get the maximum square-shaped hole is by removing horizontal bar 2 and vertical bar 2.
Example 2:
Input: n = 1, m = 1, hBars = [2], vBars = [2]
Output: 4
Explanation:
To get the maximum square-shaped hole, we remove horizontal bar 2 and vertical bar 2.
Example 3:
Input: n = 2, m = 3, hBars = [2,3], vBars = [2,4]
Output: 4
Explanation:
One way to get the maximum square-shaped hole is by removing horizontal bar 3, and vertical bar 4.
Constraints:
1 <= n <= 1091 <= m <= 1091 <= hBars.length <= 1002 <= hBars[i] <= n + 11 <= vBars.length <= 1002 <= vBars[i] <= m + 1hBars are distinct.vBars are distinct.Problem summary: You are given the two integers, n and m and two integer arrays, hBars and vBars. The grid has n + 2 horizontal and m + 2 vertical bars, creating 1 x 1 unit cells. The bars are indexed starting from 1. You can remove some of the bars in hBars from horizontal bars and some of the bars in vBars from vertical bars. Note that other bars are fixed and cannot be removed. Return an integer denoting the maximum area of a square-shaped hole in the grid, after removing some bars (possibly none).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
2 1 [2,3] [2]
1 1 [2] [2]
2 3 [2,3] [2,4]
maximal-square)maximum-square-area-by-removing-fences-from-a-field)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2943: Maximize Area of Square Hole in Grid
class Solution {
public int maximizeSquareHoleArea(int n, int m, int[] hBars, int[] vBars) {
int x = Math.min(f(hBars), f(vBars));
return x * x;
}
private int f(int[] nums) {
Arrays.sort(nums);
int ans = 1, cnt = 1;
for (int i = 1; i < nums.length; ++i) {
if (nums[i] == nums[i - 1] + 1) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 1;
}
}
return ans + 1;
}
}
// Accepted solution for LeetCode #2943: Maximize Area of Square Hole in Grid
func maximizeSquareHoleArea(n int, m int, hBars []int, vBars []int) int {
f := func(nums []int) int {
sort.Ints(nums)
ans, cnt := 1, 1
for i, x := range nums[1:] {
if x == nums[i]+1 {
cnt++
ans = max(ans, cnt)
} else {
cnt = 1
}
}
return ans + 1
}
x := min(f(hBars), f(vBars))
return x * x
}
# Accepted solution for LeetCode #2943: Maximize Area of Square Hole in Grid
class Solution:
def maximizeSquareHoleArea(
self, n: int, m: int, hBars: List[int], vBars: List[int]
) -> int:
def f(nums: List[int]) -> int:
nums.sort()
ans = cnt = 1
for i in range(1, len(nums)):
if nums[i] == nums[i - 1] + 1:
cnt += 1
ans = max(ans, cnt)
else:
cnt = 1
return ans + 1
return min(f(hBars), f(vBars)) ** 2
// Accepted solution for LeetCode #2943: Maximize Area of Square Hole in Grid
impl Solution {
pub fn maximize_square_hole_area(n: i32, m: i32, h_bars: Vec<i32>, v_bars: Vec<i32>) -> i32 {
let f = |nums: &mut Vec<i32>| -> i32 {
let mut ans = 1;
let mut cnt = 1;
nums.sort();
for i in 1..nums.len() {
if nums[i] == nums[i - 1] + 1 {
cnt += 1;
ans = ans.max(cnt);
} else {
cnt = 1;
}
}
ans + 1
};
let mut h_bars = h_bars;
let mut v_bars = v_bars;
let x = f(&mut h_bars).min(f(&mut v_bars));
x * x
}
}
// Accepted solution for LeetCode #2943: Maximize Area of Square Hole in Grid
function maximizeSquareHoleArea(n: number, m: number, hBars: number[], vBars: number[]): number {
const f = (nums: number[]): number => {
nums.sort((a, b) => a - b);
let [ans, cnt] = [1, 1];
for (let i = 1; i < nums.length; ++i) {
if (nums[i] === nums[i - 1] + 1) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 1;
}
}
return ans + 1;
};
return Math.min(f(hBars), f(vBars)) ** 2;
}
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.