Maximize Area of Square Hole in Grid

you can remove limited hbars and vbars return biggest square you can form

Maximize Area of Square Hole in Grid

Problem

You are given two integers n and m, and two arrays hBars and vBars.

• The grid has n + 2 horizontal bars and m + 2 vertical bars.
• The bars are numbered from 1.
• Together they make cells of size 1 x 1.

You can remove some bars from hBars (horizontal) and some bars from vBars (vertical). All other bars are fixed and cannot be removed.

Return the maximum area of a square-shaped hole you can make after removing bars (you may remove none).

Examples

Example 1

Input: n = 2, m = 1, hBars = [2,3], vBars = [2]
Output: 4

• The horizontal bars are [1,2,3,4] and the vertical bars are [1,2,3].
• Remove horizontal bar 2 and vertical bar 2.
• The hole is 2 x 2, so the area is 4.

Example 2

Input: n = 1, m = 1, hBars = [2], vBars = [2]
Output: 4

• Remove horizontal bar 2 and vertical bar 2.
• The hole is 2 x 2, so the area is 4.

Example 3

Input: n = 2, m = 3, hBars = [2,3], vBars = [2,4]
Output: 4

• Remove horizontal bar 3 and vertical bar 4.
• The hole is 2 x 2, so the area is 4.
• We cannot use both 2 and 4 in vBars together, because they are not next to each other.

Constraints

• 1 <= n, m <= 10^9
• 1 <= hBars.length, vBars.length <= 100
• 2 <= hBars[i] <= n + 1
• 2 <= vBars[i] <= m + 1
• All values in hBars are distinct. All values in vBars are distinct.

Main Idea

Removing bars makes a bigger gap

• Each cell is 1 x 1, so the gap between two neighbor bars is 1.
• If we remove x bars that sit next to each other (like bars 2, 3, 4), the gap between the two bars around them becomes x + 1.
• If the removed bars are not next to each other (like 2 and 4), they make separate gaps of size 2. They do not join into one bigger gap.

So only consecutive bars help us grow the hole.

A square needs equal width and height

• Let hMax be the longest run of consecutive numbers in hBars. The tallest hole we can make has height hMax + 1.
• Let vMax be the longest run of consecutive numbers in vBars. The widest hole we can make has width vMax + 1.
• For a square, both sides must be equal. We can always remove fewer bars to make a smaller side.
• So the biggest square side is min(hMax, vMax) + 1.

The answer is side * side.

Why is n and m not needed? Every bar in hBars and vBars is between 2 and n + 1 (or m + 1). The fixed outer bars 1 and n + 2 are never removed, so the gap always has a bar at each end. The values of n and m do not change the answer.

Solution: Sort and Find the Longest Streak

var maximizeSquareHoleArea = function (n, m, hBars, vBars) {
    // helper: find the longest run of consecutive numbers
    const getMaxConsecutive = (bars) => {
        bars.sort((a, b) => a - b);

        let maxStreak = 1;
        let currentStreak = 1;

        for (let i = 1; i < bars.length; i++) {
            if (bars[i] === bars[i - 1] + 1) {
                // next number in the run
                currentStreak++;
            } else {
                // run is broken, start again
                currentStreak = 1;
            }

            maxStreak = Math.max(maxStreak, currentStreak);
        }

        return maxStreak;
    };

    const hMax = getMaxConsecutive(hBars);
    const vMax = getMaxConsecutive(vBars);

    // the side is (removed bars + 1), and both sides must be equal
    const side = Math.min(hMax, vMax) + 1;

    return side * side;
};

How the Code Works

Step 1: Sort the bars

• bars.sort((a, b) => a - b) puts the numbers in increasing order.
• We pass (a, b) => a - b, because the default JavaScript sort compares numbers as text. For example, 10 would come before 2.
• After sorting, consecutive numbers sit next to each other.

Step 2: Find the longest streak

• currentStreak is the length of the run we are in now. maxStreak is the best run so far.
• Both start at 1, because a single bar is a run of length 1.
• If bars[i] === bars[i - 1] + 1, the run continues, so we add 1.
• Otherwise, the run is broken, and we reset to 1.

Step 3: Combine both directions

• hMax and vMax come from the same helper.
• Math.min(hMax, vMax) + 1 is the biggest square side.
• We return side * side.

Dry Run

Let us take hBars = [2, 3] and vBars = [2, 4].

Horizontal bars (after sorting [2, 3]):

So hMax = 2.

Vertical bars (after sorting [2, 4]):

So vMax = 1.

Result:

• side = min(2, 1) + 1 = 2
• Area = 2 * 2 = 4

This matches the expected output.

Complexity

• Time Complexity: O(h log h + v log v)
  • h is the length of hBars and v is the length of vBars.
  • Sorting is the biggest cost. The streak loop takes O(h + v).
  • Both arrays have at most 100 items, so it is very fast.
• Space Complexity: O(1)
  • We use only a few variables. Sorting happens in place.

Note: sort changes the input arrays. This is fine here. If you must keep the original arrays, sort a copy with [...bars].sort(...).

Comparison

Summary

• Removing x consecutive bars makes a gap of size x + 1.
• Bars that are not next to each other do not join into one bigger gap.
• So we need the longest run of consecutive numbers in hBars and in vBars.
• A square needs equal sides, so the side is min(hMax, vMax) + 1.
• The answer is side * side.
• n and m are not used, because the fixed outer bars always give the gap its two ends.
• The solution runs in O(h log h + v log v) time and uses O(1) space.