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You are given an integer n that indicates the size of a n x n binary grid. Initially, every cell in the grid is a 1. You are also given an array mines where mines[i] = [xi, yi] indicates that the cell (xi, yi) receives a 0. This action turns the cell (xi, yi) into a mine.

You need to find the order of the largest axis-aligned plus sign of 1s contained in the grid after performing these operations. An order of the largest plus sign is defined as the largest integer k such that there exists a plus sign of order k with its center in some cell of the grid, which contains only 1s. A plus sign of order k has a center cell (r, c) with the following property that must be true for all 0 <= i < k:

Return an integer that represents the order of the largest plus sign.

Clarifying Questions

  1. Input Constraints:
    • What is the range of n?
    • How many elements can the mines array have?
  2. Output Requirements:
    • Should we return 0 if there is no valid plus sign?
  3. Edge Cases:
    • What to do when the entire grid is filled with mines?
    • What if mines is empty and there are no 0s in the grid?

Strategy

  1. Initialization:
    • Create a grid initialized with 1s.
    • Set cells to 0 as per the mines array.
  2. Pre-computation:
    • Use four auxiliary arrays to keep track of the number of continuous 1s in all four possible directions from any cell:
      • Left to right
      • Right to left
      • Top to bottom
      • Bottom to top
  3. Cell-wise Computation:
    • For each cell (i, j), compute the minimum value from the four auxiliary arrays, which gives the maximum k for the plus sign centered at (i, j).
  4. Result Preparation:
    • The final result will be the maximum value found in the above step.

Code

def orderOfLargestPlusSign(n, mines):
    if n == 0:
        return 0

    # Initialize the grid
    grid = [[1] * n for _ in range(n)]
    for mine in mines:
        grid[mine[0]][mine[1]] = 0

    left = [[0] * n for _ in range(n)]
    right = [[0] * n for _ in range(n)]
    top = [[0] * n for _ in range(n)]
    bottom = [[0] * n for _ in range(n)]
    
    for i in range(n):
        for j in range(n):
            if grid[i][j] == 1:
                left[i][j] = left[i][j-1] + 1 if j > 0 else 1
                top[i][j] = top[i-1][j] + 1 if i > 0 else 1

    for i in range(n-1, -1, -1):
        for j in range(n-1, -1, -1):
            if grid[i][j] == 1:
                right[i][j] = right[i][j+1] + 1 if j < n-1 else 1
                bottom[i][j] = bottom[i+1][j] + 1 if i < n-1 else 1

    max_order = 0
    for i in range(n):
        for j in range(n):
            if grid[i][j] == 1:
                order = min(left[i][j], right[i][j], top[i][j], bottom[i][j])
                max_order = max(max_order, order)

    return max_order

Time Complexity

Overall, the time complexity is O(n^2). This is efficient given that we need to consider every cell in the grid and the computations for each cell are constant time operations.

Edge Cases

  1. If n is 0, the function returns 0 as there’s no grid.
  2. If mines covers the entire grid, the function returns 0 since no 1-cell exists.
  3. If mines is empty, the order of the largest plus sign will be n//2 + 1.

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