In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, return 0.
An "axis-aligned plus sign of 1s of order k" has some center grid[x][y] = 1 along with 4 arms of length k-1 going up, down, left, and right, and made of 1s. This is demonstrated in the diagrams below. Note that there could be 0s or 1s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1s.
Examples of Axis-Aligned Plus Signs of Order k:
Order 1: 000 010 000 Order 2: 00000 00100 01110 00100 00000 Order 3: 0000000 0001000 0001000 0111110 0001000 0001000 0000000
Example 1:
Input: N = 5, mines = [[4, 2]] Output: 2 Explanation: 11111 11111 11111 11111 11011 In the above grid, the largest plus sign can only be order 2. One of them is marked in bold.
Example 2:
Input: N = 2, mines = [] Output: 1 Explanation: There is no plus sign of order 2, but there is of order 1.
Example 3:
Input: N = 1, mines = [[0, 0]] Output: 0 Explanation: There is no plus sign, so return 0.
Note:
N will be an integer in the range [1, 500].mines will have length at most 5000.mines[i] will be length 2 and consist of integers in the range [0, N-1].Intuition and Algorithm
\nFor each possible center, find the largest plus sign that could be placed by repeatedly expanding it.\nWe expect this algorithm to be , and so take roughly operations. This is a little bit too big for us to expect it to run in time.
\n\nComplexity Analysis
\nTime Complexity: , as we perform two outer loops (), plus the inner loop involving k is .
Space Complexity: .
\nIntuition
\nHow can we improve our bruteforce? One way is to try to speed up the inner loop involving k, the order of the candidate plus sign.\nIf we knew the longest possible arm length in each direction from a center, we could know the order of a plus sign at that center. We could find these lengths separately using dynamic programming.
Algorithm
\nFor each (cardinal) direction, and for each coordinate (r, c) let\'s compute the count of that coordinate: the longest line of \'1\'s starting from (r, c) and going in that direction.\nWith dynamic programming, it is either 0 if grid[r][c] is zero, else it is 1 plus the count of the coordinate in the same direction.\nFor example, if the direction is left and we have a row like 01110110, the corresponding count values are 01230120, and the integers are either 1 more than their successor, or 0.\nFor each square, we want dp[r][c] to end up being the minimum of the 4 possible counts. At the end, we take the maximum value in dp.
Complexity Analysis
\nTime Complexity: , as the work we do under two nested for loops is .
\nSpace Complexity: , the size of dp.
Analysis written by: @awice.
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