764. Largest Plus Sign

Problem:

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.

Analysis:
The idea is similar to 221. Maximal Square. The current plus sign is determined by the smallest length on each direction. So we use dp to calc length of each direction on every point of the matrix.

Solution:

class Solution {
    public int orderOfLargestPlusSign(int N, int[][] mines) {
        Set<Integer> set = new HashSet<>();
        int[][] dp = new int[N][N];
        int res = 0, count = 0;

        for (int i = 0; i < mines.length; i++) {
            set.add(mines[i][0]*N + mines[i][1]);
        }

        for (int j = 0; j < N; j++) {
            count = 0;
            for (int i = 0; i < N; i++) { // up
                count = set.contains(i * N + j) ? 0 : count + 1;
                dp[i][j] = count;
            }
            count = 0;
            for (int i = N - 1; i >= 0; i--) { // down
                count = set.contains(i * N + j) ? 0 : count + 1;
                dp[i][j] = Math.min(count, dp[i][j]);
            }
        }

        for (int i = 0; i < N; i++) {
            count = 0;
            for (int j = 0; j < N; j++) { // left
                count = set.contains(i * N + j) ? 0 : count + 1;
                dp[i][j] = Math.min(count, dp[i][j]);
            }
            count = 0;
            for (int j = N - 1; j >= 0; j--) { // right
                count = set.contains(i * N + j) ? 0 : count + 1;
                dp[i][j] = Math.min(count, dp[i][j]);
                res = Math.max(res, dp[i][j]);
            }
        }

        return res;
    }
}