🧱 Beginner-Friendly Guide 'Maximize Area of Square Hole in Grid' – LeetCode 2943 (C++, Python, JavaScript)

Understanding how to manipulate grids and find contiguous patterns is a vital skill for any developer. This problem challenges us to think about how removing physical constraints (the bars) can create a larger unified space (the hole).

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Problem Summary

You're given:

• Two integers and , representing the number of internal horizontal and vertical bars.
• Two arrays, hBars and vBars, which list the specific bars that are eligible for removal.

Your goal:

• Find the maximum area of a square hole you can create by removing any number of the removable bars.

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Intuition

The grid is made of cells separated by bars. To make a hole larger, we must remove consecutive bars. If we remove one bar, the gap between the two remaining fixed bars becomes 2 units. If we remove two consecutive bars, the gap becomes 3 units.

Essentially, the length of a "hole" in one direction is equal to the number of consecutive bars removed plus 1.

1. Find the Maximum Gap: For both horizontal and vertical directions, we sort the removable bars and look for the longest sequence of consecutive numbers (e.g., bars 2, 3, and 4).
2. Consecutive Logic: If we find consecutive bars in hBars, the maximum horizontal gap we can form is .
3. The Square Constraint: Since the hole must be a square, its side length is limited by the smaller of the two maximum gaps (horizontal vs. vertical). If the max horizontal gap is 5 and the max vertical gap is 3, the largest square we can fit is a .
4. Final Calculation: Once we find the maximum possible side length , the area is simply .

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Code Blocks

C++

class Solution {
public:
    int maximizeSquareHoleArea(int n, int m, vector<int>& hBars, vector<int>& vBars) {
        // The maximum square side is the minimum of the best horizontal and vertical gaps
        int side = min(getMaxGap(hBars), getMaxGap(vBars));
        return side * side;
    }

private:
    int getMaxGap(vector<int>& bars) {
        int max_consecutive = 1;
        int current_consecutive = 1;
        sort(bars.begin(), bars.end());

        for (int i = 1; i < bars.size(); ++i) {
            if (bars[i] == bars[i - 1] + 1) {
                current_consecutive++;
            } else {
                current_consecutive = 1;
            }
            max_consecutive = max(max_consecutive, current_consecutive);
        }
        // A sequence of 'k' bars removed creates a gap of 'k + 1'
        return max_consecutive + 1;
    }
};

Python

class Solution:
    def maximizeSquareHoleArea(self, n: int, m: int, hBars: list[int], vBars: list[int]) -> int:
        def get_max_gap(bars: list[int]) -> int:
            bars.sort()
            max_consecutive = 1
            current_consecutive = 1

            for i in range(1, len(bars)):
                if bars[i] == bars[i - 1] + 1:
                    current_consecutive += 1
                else:
                    current_consecutive = 1
                max_consecutive = max(max_consecutive, current_consecutive)

            # The gap width is the number of consecutive bars + 1
            return max_consecutive + 1

        side = min(get_max_gap(hBars), get_max_gap(vBars))
        return side * side

JavaScript

/**
 * @param {number} n
 * @param {number} m
 * @param {number} hBars
 * @param {number} vBars
 * @return {number}
 */
var maximizeSquareHoleArea = function(n, m, hBars, vBars) {
    const getMaxGap = (bars) => {
        bars.sort((a, b) => a - b);
        let maxConsecutive = 1;
        let currentConsecutive = 1;

        for (let i = 1; i < bars.length; i++) {
            if (bars[i] === bars[i - 1] + 1) {
                currentConsecutive++;
            } else {
                currentConsecutive = 1;
            }
            maxConsecutive = Math.max(maxConsecutive, currentConsecutive);
        }
        return maxConsecutive + 1;
    };

    const side = Math.min(getMaxGap(hBars), getMaxGap(vBars));
    return side * side;
};

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Key Takeaways

• Sorting for Contiguity: When looking for consecutive sequences in a dataset, sorting is usually the most efficient first step, bringing the time complexity but simplifying the logic to a single linear pass.
• The "Fencepost" Error: Always remember that bars separate spaces. In this problem, removing consecutive bars merges units of space.
• Constraint Bottlenecks: In geometry problems involving squares, the smallest dimension (the bottleneck) always dictates the final size.

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Final Thoughts

This problem is an excellent example of how to simplify a 2D grid problem into two 1D sub-problems. In real-world software engineering, this "bottleneck" logic is used frequently in resource allocation and window management systems, where a viewing area is restricted by the most limited dimension available. Mastering these types of problems helps you develop an eye for identifying which constraints truly matter in a complex system.