Introduction

In this blog post, we will explore how to solve the LeetCode problem "2943" 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

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 <= 109 1 <= m <= 109 1 <= hBars.length <= 100 2 <= hBars[i] <= n + 1 1 <= vBars.length <= 100 2 <= vBars[i] <= m + 1 All values in hBars are distinct. All values in vBars are distinct.

Explanation

Here's a breakdown of the solution:

• Find Maximum Consecutive Horizontal Gaps: Determine the largest number of consecutive horizontal bars that can be removed. This is done by sorting the hBars and calculating the maximum difference between consecutive elements after adding the boundary bars (1 and n+2).
• Find Maximum Consecutive Vertical Gaps: Similarly, find the largest number of consecutive vertical bars that can be removed by processing vBars along with boundary bars (1 and m+2).

• Determine Maximum Square Hole: The maximum size of the square hole is the smaller of the maximum horizontal gap and the maximum vertical gap. The area is the square of this value.

• Runtime Complexity: O(h log h + v log v), where h is the number of elements in hBars and v is the number of elements in vBars (due to sorting). Storage Complexity: O(1) excluding input arrays.

Code

    def solve():
    n, m, hBars, vBars = map(eval, input().splitlines())

    def max_consecutive_gap(bars, limit):
        bars.sort()
        max_gap = 0
        prev = 1
        for bar in bars:
            max_gap = max(max_gap, bar - prev)
            prev = bar + 1
        max_gap = max(max_gap, limit + 2 - prev)
        return max_gap

    horizontal_gap = max_consecutive_gap(hBars, n)
    vertical_gap = max_consecutive_gap(vBars, m)

    side = min(horizontal_gap, vertical_gap)
    return side * side

def main():
    print(solve())

if __name__ == "__main__":
    import sys
    input = sys.stdin.readline  # for faster I/O

    # For multiple test cases (if needed):
    # t = int(input())
    # for _ in range(t):
    #     print(solve())

    # For single test case:
    main()