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 1
s contained in the grid? Return the order of the plus sign. If there is none, return 0.
An "axis-aligned plus sign of
1
s 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 1
s. This is demonstrated in the diagrams below. Note that there could be 0
s or 1
s 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 most5000
.mines[i]
will be length 2 and consist of integers in the range[0, N-1]
.- (Additionally, programs submitted in C, C++, or C# will be judged with a slightly smaller time limit.)
class Solution { public int orderOfLargestPlusSign(int N, int[][] mines) { Set<Integer> set = new HashSet<>(); for (int[] pair : mines) { set.add(pair[0] * N + pair[1]); } int[][] dp = new int[N][N]; int ans = 0; for (int i = 0; i < N; i++) { int count = 0; // left // for (int j = 0; j < N; j++) { count = set.contains(i * N + j) ? 0 : count + 1; dp[i][j] = count; } // right // count = 0; for (int j = N - 1; j >= 0; j--) { count = set.contains(i * N + j) ? 0 : count + 1; dp[i][j] = Math.min(dp[i][j], count); } } for (int j = 0; j < N; j++) { int count = 0; // up // for (int i = 0; i < N; i++) { count = set.contains(i * N + j) ? 0 : count + 1; dp[i][j] = Math.min(dp[i][j], count); } count = 0; // down // for (int i = N - 1; i >= 0; i--) { count = set.contains(i * N + j) ? 0 : count + 1; dp[i][j] = Math.min(dp[i][j], count); ans = Math.max(ans,dp[i][j]); } } return ans; } }
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