{ "data": { "question": { "questionId": "3214", "questionFrontendId": "2943", "boundTopicId": null, "title": "Maximize Area of Square Hole in Grid", "titleSlug": "maximize-area-of-square-hole-in-grid", "content": "

There is a grid with n + 2 horizontal bars and m + 2 vertical bars, and initially containing 1 x 1 unit cells.

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The bars are 1-indexed.

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You are given the two integers, n and m.

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You are also given two integer arrays: hBars and vBars.

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You are allowed to remove bars that satisfy any of the following conditions:

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Return an integer denoting the maximum area of a square-shaped hole in the grid after removing some bars (possibly none).

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Example 1:

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\nInput: n = 2, m = 1, hBars = [2,3], vBars = [2]\nOutput: 4\nExplanation: The left image shows the initial grid formed by the bars.\nThe horizontal bars are in the range [1,4], and the vertical bars are in the range [1,3].\nIt is allowed to remove horizontal bars [2,3] and the vertical bar [2].\nOne way to get the maximum square-shaped hole is by removing horizontal bar 2 and vertical bar 2.\nThe resulting grid is shown on the right.\nThe hole has an area of 4.\nIt can be shown that it is not possible to get a square hole with an area more than 4.\nHence, the answer is 4.\n
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Example 2:

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\"\"

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\nInput: n = 1, m = 1, hBars = [2], vBars = [2]\nOutput: 4\nExplanation: The left image shows the initial grid formed by the bars.\nThe horizontal bars are in the range [1,3], and the vertical bars are in the range [1,3].\nIt is allowed to remove the horizontal bar [2] and the vertical bar [2].\nTo get the maximum square-shaped hole, we remove horizontal bar 2 and vertical bar 2.\nThe resulting grid is shown on the right.\nThe hole has an area of 4.\nHence, the answer is 4, and it is the maximum possible.\n
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Example 3:

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\"\"

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\nInput: n = 2, m = 3, hBars = [2,3], vBars = [2,3,4]\nOutput: 9\nExplanation: The left image shows the initial grid formed by the bars.\nThe horizontal bars are in the range [1,4], and the vertical bars are in the range [1,5].\nIt is allowed to remove horizontal bars [2,3] and vertical bars [2,3,4].\nOne way to get the maximum square-shaped hole is by removing horizontal bars 2 and 3, and vertical bars 3 and 4.\nThe resulting grid is shown on the right.\nThe hole has an area of 9.\nIt can be shown that it is not possible to get a square hole with an area more than 9.\nHence, the answer is 9.\n
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Constraints:

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Compiled with clang 11 using the latest C++ 20 standard.

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Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.

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Most standard library headers are already included automatically for your convenience.

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Python 2.7.12.

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