2975. Maximum Square Area by Removing Fences From a Field

Description

There is a large (m - 1) x (n - 1) rectangular field with corners at (1, 1) and (m, n) containing some horizontal and vertical fences given in arrays hFences and vFences respectively.

Horizontal fences are from the coordinates (hFences[i], 1) to (hFences[i], n) and vertical fences are from the coordinates (1, vFences[i]) to (m, vFences[i]).

Return the maximum area of a square field that can be formed by removing some fences (possibly none) or -1 if it is impossible to make a square field.

Since the answer may be large, return it modulo 109 + 7.

Note: The field is surrounded by two horizontal fences from the coordinates (1, 1) to (1, n) and (m, 1) to (m, n) and two vertical fences from the coordinates (1, 1) to (m, 1) and (1, n) to (m, n). These fences cannot be removed.

 

Example 1:

Input: m = 4, n = 3, hFences = [2,3], vFences = [2]
Output: 4
Explanation: Removing the horizontal fence at 2 and the vertical fence at 2 will give a square field of area 4.

Example 2:

Input: m = 6, n = 7, hFences = [2], vFences = [4]
Output: -1
Explanation: It can be proved that there is no way to create a square field by removing fences.

 

Constraints:

  • 3 <= m, n <= 109
  • 1 <= hFences.length, vFences.length <= 600
  • 1 < hFences[i] < m
  • 1 < vFences[i] < n
  • hFences and vFences are unique.

Solutions

Solution 1: Enumeration

We can enumerate any two horizontal fences $a$ and $b$ in $hFences$, calculate the distance $d$ between $a$ and $b$, and record it in the hash table $hs$. Then, we enumerate any two vertical fences $c$ and $d$ in $vFences$, calculate the distance $d$ between $c$ and $d$, and record it in the hash table $vs$. Finally, we traverse the hash table $hs$. If a distance $d$ in $hs$ also exists in the hash table $vs$, it means that there exists a square field with a side length of $d$ and an area of $d^2$. We just need to take the largest $d$ and calculate $d^2 \bmod 10^9 + 7$.

The time complexity is $O(h^2 + v^2)$, and the space complexity is $O(h^2 + v^2)$. Where $h$ and $v$ are the lengths of $hFences$ and $vFences$ respectively.

Python Code
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class Solution:
    def maximizeSquareArea(
        self, m: int, n: int, hFences: List[int], vFences: List[int]
    ) -> int:
        def f(nums: List[int], k: int) -> Set[int]:
            nums.extend([1, k])
            nums.sort()
            return {b - a for a, b in combinations(nums, 2)}

        mod = 10**9 + 7
        hs = f(hFences, m)
        vs = f(vFences, n)
        ans = max(hs & vs, default=0)
        return ans**2 % mod if ans else -1

Java Code
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class Solution {
    public int maximizeSquareArea(int m, int n, int[] hFences, int[] vFences) {
        Set<Integer> hs = f(hFences, m);
        Set<Integer> vs = f(vFences, n);
        hs.retainAll(vs);
        int ans = -1;
        final int mod = (int) 1e9 + 7;
        for (int x : hs) {
            ans = Math.max(ans, x);
        }
        return ans > 0 ? (int) (1L * ans * ans % mod) : -1;
    }

    private Set<Integer> f(int[] nums, int k) {
        int n = nums.length;
        nums = Arrays.copyOf(nums, n + 2);
        nums[n] = 1;
        nums[n + 1] = k;
        Arrays.sort(nums);
        Set<Integer> s = new HashSet<>();
        for (int i = 0; i < nums.length; ++i) {
            for (int j = 0; j < i; ++j) {
                s.add(nums[i] - nums[j]);
            }
        }
        return s;
    }
}

C++ Code
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class Solution {
public:
    int maximizeSquareArea(int m, int n, vector<int>& hFences, vector<int>& vFences) {
        auto f = [](vector<int>& nums, int k) {
            nums.push_back(k);
            nums.push_back(1);
            sort(nums.begin(), nums.end());
            unordered_set<int> s;
            for (int i = 0; i < nums.size(); ++i) {
                for (int j = 0; j < i; ++j) {
                    s.insert(nums[i] - nums[j]);
                }
            }
            return s;
        };
        auto hs = f(hFences, m);
        auto vs = f(vFences, n);
        int ans = 0;
        for (int h : hs) {
            if (vs.count(h)) {
                ans = max(ans, h);
            }
        }
        const int mod = 1e9 + 7;
        return ans > 0 ? 1LL * ans * ans % mod : -1;
    }
};

Go Code
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func maximizeSquareArea(m int, n int, hFences []int, vFences []int) int {
	f := func(nums []int, k int) map[int]bool {
		nums = append(nums, 1, k)
		sort.Ints(nums)
		s := map[int]bool{}
		for i := 0; i < len(nums); i++ {
			for j := 0; j < i; j++ {
				s[nums[i]-nums[j]] = true
			}
		}
		return s
	}
	hs := f(hFences, m)
	vs := f(vFences, n)
	ans := 0
	for h := range hs {
		if vs[h] {
			ans = max(ans, h)
		}
	}
	if ans > 0 {
		return ans * ans % (1e9 + 7)
	}
	return -1
}

TypeScript Code
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function maximizeSquareArea(m: number, n: number, hFences: number[], vFences: number[]): number {
    const f = (nums: number[], k: number): Set<number> => {
        nums.push(1, k);
        nums.sort((a, b) => a - b);
        const s: Set<number> = new Set();
        for (let i = 0; i < nums.length; ++i) {
            for (let j = 0; j < i; ++j) {
                s.add(nums[i] - nums[j]);
            }
        }
        return s;
    };
    const hs = f(hFences, m);
    const vs = f(vFences, n);
    let ans = 0;
    for (const h of hs) {
        if (vs.has(h)) {
            ans = Math.max(ans, h);
        }
    }
    return ans ? Number(BigInt(ans) ** 2n % 1000000007n) : -1;
}