Introduction

In this blog post, we will explore how to solve the LeetCode problem "2975" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the solution approach, complexities, and the code:

• Key Idea: The problem requires finding the largest possible square formed by existing fences. This can be achieved by pre-processing fence locations to compute all possible distances between them both horizontally and vertically. Then check for equal horizontal and vertical distances, and return the square of the largest matching distance modulo 10⁹ + 7.
• Optimization: Using sets for distance lookups provides fast checks for corresponding horizontal and vertical fence gaps, making the algorithm efficient.

• Edge Case: Need to include the boundaries 1 and m (horizontal) and 1 and n (vertical) as fences.

• Complexity:

  • Runtime: O(h² + v²), where h is the number of horizontal fences and v is the number of vertical fences.
  • Storage: O(h² + v²)

Code

    def maximizeSquareArea(m: int, n: int, hFences: list[int], vFences: list[int]) -> int:
    """
    Finds the maximum area of a square field formed by removing fences.

    Args:
        m: The upper bound of the horizontal coordinate (m).
        n: The upper bound of the vertical coordinate (n).
        hFences: A list of horizontal fence coordinates.
        vFences: A list of vertical fence coordinates.

    Returns:
        The maximum area of a square field modulo 10^9 + 7, or -1 if no square can be formed.
    """

    hFences = sorted(hFences + [1, m])
    vFences = sorted(vFences + [1, n])

    h_diffs = set()
    for i in range(len(hFences)):
        for j in range(i + 1, len(hFences)):
            h_diffs.add(hFences[j] - hFences[i])

    v_diffs = set()
    for i in range(len(vFences)):
        for j in range(i + 1, len(vFences)):
            v_diffs.add(vFences[j] - vFences[i])

    max_side = 0
    for side in h_diffs:
        if side in v_diffs:
            max_side = max(max_side, side)

    if max_side == 0:
        return -1
    else:
        return (max_side * max_side) % (10**9 + 7)