Description
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 + 1- All values in
hBarsare distinct. - All values in
vBarsare distinct.
Solutions
This solution finds the largest square hole in a grid by identifying the longest consecutive sequences of missing horizontal and vertical bars. It first sorts both arrays, then iterates through each to count consecutive integers (where each bar is exactly 1 position after the previous one), tracking the maximum consecutive count for both dimensions. The side length of the largest possible square is determined by taking the Math.min() of these two maxima and adding 1 (since a sequence of length n creates a hole of size n+1), and the function returns the area by squaring this side length.
/**
* @param {number} n
* @param {number} m
* @param {number[]} hBars
* @param {number[]} vBars
* @return {number}
*/
var maximizeSquareHoleArea = function (n, m, hBars, vBars) {
hBars.sort((a, b) => a - b);
vBars.sort((a, b) => a - b);
let hmax = 1;
let vmax = 1;
let hcur = 1;
let vcur = 1;
for (let i = 1; i < hBars.length; i++) {
if (hBars[i] === hBars[i - 1] + 1) {
hcur++;
} else {
hcur = 1;
}
hmax = Math.max(hmax, hcur);
}
for (let i = 1; i < vBars.length; i++) {
if (vBars[i] === vBars[i - 1] + 1) {
vcur++;
} else {
vcur = 1;
}
vmax = Math.max(vmax, vcur);
}
const side = Math.min(hmax, vmax) + 1;
return side * side;
};